Using Gram-Schmidt orthogonalization, find an orthogonal basis for the span of the vectors $\mathbf{w}_{1},\mathbf{w}_{2}\in\R^{3}$ if
\[
\mathbf{w}_{1}
=
\begin{bmatrix}
1 \\ 0 \\ 3
\end{bmatrix}
,\quad
\mathbf{w}_{2}
=
\begin{bmatrix}
2 \\ -1 \\ 0
\end{bmatrix}
.
\]

Solution.

We apply Gram-Schmidt orthogonalization as follows. The first step is to define $\mathbf{u}_{1}=\mathbf{w}_{1}$. Before defining $\mathbf{u}_{2}$, we must compute
\begin{align*}
\mathbf{u}_{1}^{T}\mathbf{w}_{2}
&=
\mathbf{w}_{1}^{T}\mathbf{w}_{2}
=
\begin{bmatrix}
1 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
2 \\ -1 \\ 0
\end{bmatrix}
=2+0+0=2,
\\
\mathbf{u}_{1}^{T}\mathbf{u}_{1}
&=
\mathbf{w}_{1}^{T}\mathbf{w}_{1}
=
\begin{bmatrix}
1 & 0 & 3
\end{bmatrix}
\begin{bmatrix}
1 \\ 0 \\ 3
\end{bmatrix}
=1+0+9=10.
\end{align*}

Next, we define
\[
\mathbf{u}_{2}
=
\mathbf{w}_{2}
-\dfrac{\mathbf{u}_{1}^{T}\mathbf{w}_{2}}
{\mathbf{u}_{1}^{T}\mathbf{u}_{1}}
\mathbf{u}_{1}
=
\begin{bmatrix}
2 \\ -1 \\ 0
\end{bmatrix}
-\dfrac{2}{10}
\begin{bmatrix}
1 \\ 0 \\ 3
\end{bmatrix}
=
\begin{bmatrix}
10/5 \\ -1 \\ 0
\end{bmatrix}
–
\begin{bmatrix}
1/5 \\ 0 \\ 3/5
\end{bmatrix}
=
\begin{bmatrix}
9/5 \\ -1 \\ -3/5
\end{bmatrix}
.
\] By Gram-Schmidt orthogonalization, $\{\mathbf{u}_{1},\mathbf{u}_{2}\}$ is an orthogonal basis for the span of the vectors $\mathbf{w}_{1}$ and $\mathbf{w}_{2}$.

Remark

Note that since scalar multiplication by a nonzero number does not change the orthogonality of vectors and the new vectors still form a basis, we could have used $5\mathbf{u}_2$, instead of $\mathbf{u}_2$ to avoid a fraction in our computation.
We have
\[5\mathbf{u}_2=\begin{bmatrix}
10 \\
-5 \\
0
\end{bmatrix}-\begin{bmatrix}
1 \\
0 \\
3
\end{bmatrix}=\begin{bmatrix}
9 \\
-5 \\
-3
\end{bmatrix},\] and $\{\mathbf{u}_1, 5\mathbf{u}_2\}$ is an orthogonal basis for the span.

The post Using Gram-Schmidt Orthogonalization, Find an Orthogonal Basis for the Span first appeared on Problems in Mathematics.

Let
\[
\mathbf{v}_{1}
=
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
,\;
\mathbf{v}_{2}
=
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
.
\] Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?

If not, then find an orthonormal basis for $V$.

Solution.

We begin by computing
\begin{align*}
\mathbf{v}_{1}\cdot\mathbf{v}_{2}
&=
\mathbf{v}_{1}^{T}\mathbf{v}_{2}
=
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
=
1\cdot 1+1\cdot -1
=
1-1
=
0,
\\
\mathbf{v}_{1}\cdot\mathbf{v}_{1}
&=
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
=
1+1
=
2,
\\
\mathbf{v}_{2}\cdot\mathbf{v}_{2}
&=
\begin{bmatrix}
1 & -1
\end{bmatrix}
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
=
1+1
=
2.
\end{align*}
Since $\mathbf{v}_{1}\cdot\mathbf{v}_{2}=0$, the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are orthogonal. Since $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are nonzero orthogonal vectors, they are linearly independent, and it follows that $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthogonal basis for $V$. However, since $\mathbf{v}_{i}\cdot\mathbf{v}_{i}=2\neq 1$ for $i=1,2$, we know that $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ do not form an orthonormal basis for $V$.

To find an orthonormal basis for $V$, note that for any scalars $a$ and $b$, $(a\mathbf{v}_{1})\cdot(b\mathbf{v}_{2})=ab(\mathbf{v}_{1}\cdot\mathbf{v}_{2})=ab\cdot 0=0$. Therefore, $a\mathbf{v}_{1}$ and $b\mathbf{v}_{2}$ will always form an orthogonal basis for $V$. All we need to do is choose $a$ and $b$ so that $a\mathbf{v}_{1}$ and $b\mathbf{v}_{2}$ form an orthonormal set. For $a$, we require
\[
1
=\|a\mathbf{v}\|=a\|\mathbf{v}\|
\] and so
\[
a=\frac{1}{\|\mathbf{v}\|}=\frac{1}{\sqrt{2}}.
\] (Note that $\|\mathbf{v}\| \neq 0$ as $\mathbf{v}\neq \mathbf{0}$. Also, note that to obtain a length 1 vector, we just needed divide the vector by its length.)
Similarly, $b=1/\sqrt{2}$. Therefore, if we define
\begin{align*}
\mathbf{w}_{1}
&=
\dfrac{1}{\sqrt{2}}
\mathbf{v}_{1}
=
\dfrac{1}{\sqrt{2}}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
,
\\
\mathbf{w}_{2}
&=
\dfrac{1}{\sqrt{2}}
\mathbf{v}_{2}
=
\dfrac{1}{\sqrt{2}}
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
,
\end{align*}
then $\mathbf{w}_{1}$ and $\mathbf{w}_{2}$ form an orthonormal basis for $V$.

The post Normalize Lengths to Obtain an Orthonormal Basis first appeared on Problems in Mathematics.

Determine bases for $\calN(A)$ and $\calN(A^{T}A)$ when
\[
A=
\begin{bmatrix}
1 & 2 & 1 \\
1 & 1 & 3 \\
0 & 0 & 0
\end{bmatrix}
.
\] Then, determine the ranks and nullities of the matrices $A$ and $A^{\trans}A$.

Solution.

We will first compute
\begin{align*}
A^{T}
=&
\begin{bmatrix}
1 & 1 & 0 \\
2 & 1 & 0 \\
1 & 3 & 0
\end{bmatrix}
\;\text{and}\\
A^{T}A
=&
\begin{bmatrix}
1 & 1 & 0 \\
2 & 1 & 0 \\
1 & 3 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 1 \\
1 & 1 & 3 \\
0 & 0 & 0
\end{bmatrix}
=
\begin{bmatrix}
1+1 & 2+1 & 1+3 \\
2+1 & 4+1 & 2+3 \\
1+3 & 2+3 & 1+9
\end{bmatrix}
=
\begin{bmatrix}
2 & 3 & 4 \\
3 & 5 & 5 \\
4 & 5 & 10
\end{bmatrix}
.
\end{align*}

Next, we will find $\calN(A)$ by row reducing $[A\mid\mathbf{0}]$:
\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
1 & 1 & 3 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] \xrightarrow{R_{2}-R_{1}}
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & -1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] \xrightarrow{-R_{2}}
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & 1 & -2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] \] \[
\xrightarrow{R_{1}-2R_{2}}
\left[\begin{array}{ccc|c}
1 & 0 & 5 & 0 \\
0 & 1 & -2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] .
\] Thus the solution to $A\mathbf{x}=\mathbf{0}$ is given by
\[
\mathbf{x}
=
\begin{bmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{bmatrix}
=
\begin{bmatrix}
-5x_{3} \\ 2x_{3} \\ x_{3}
\end{bmatrix}
=
x_{3}
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix}
.
\] Thus
\begin{align*}
\calN(A)
=
\left\{\mathbf{x}\in \R^3 \quad\middle|\quad \mathbf{x}=x_{3}
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix} \text{ for any } x_{3}\in\R
\right\}
=\Span\left\{
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix}
\right\}
.
\end{align*}
Therefore,
\[
\left\{
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix}
\right\}
\] is a basis for $\calN(A)$.

Similarly, we will compute $\calN(A^{T}A)$ by row reducing $[A^{T}A\mid\mathbf{0}]$:
\[
\left[\begin{array}{ccc|c}
2 & 3 & 4 & 0 \\
3 & 5 & 5 & 0 \\
4 & 5 & 10 & 0
\end{array}\right] \xrightarrow[R_{3}-2R_{1}]{R_{2}-R_{1}}
\left[\begin{array}{ccc|c}
2 & 3 & 4 & 0 \\
1 & 2 & 1 & 0 \\
0 & -1 & 2 & 0
\end{array}\right] \xrightarrow{R_{1}-2R_{2}}
\left[\begin{array}{ccc|c}
0 & -1 & 2 & 0 \\
1 & 2 & 1 & 0 \\
0 & -1 & 2 & 0
\end{array}\right] \] \[
\xrightarrow{R_{3}-R_{1}}
\left[\begin{array}{ccc|c}
0 & -1 & 2 & 0 \\
1 & 2 & 1 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] \xrightarrow{R_{1}\leftrightarrow R_{2}}
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & -1 & 2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] \] \[
\xrightarrow[\text{then}\;-R_{2}]{R_{1}+2R_{2}}
\left[\begin{array}{ccc|c}
1 & 0 & 5 & 0 \\
0 & 1 & -2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right] .
\] Since the row reduced matrices for $[A\mid\mathbf{0}]$ and $[A^{T}A\mid\mathbf{0}]$ are identical, we can immediately conclude that
\[
\calN\left(A^{T}A\right)
=
\Span\left\{
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix}
\right\}
,
\] and that
\[
\left\{
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix}
\right\}
\] is a basis for $\calN(A^{T}A)$.

It follows that the nullities of $A$ and $A^{\trans}A$ are both $1$.
The rank-nullity theorem tells
\[\text{rank of $A$} + \text{nullity of $A$} =3.\] Hence the rank of $A$ is $2$. Similarly, the rank of $A^{\trans}A$ is $2$.

The post Determine Bases for Nullspaces $\calN(A)$ and $\calN(A^{T}A)$ first appeared on Problems in Mathematics.

Find a basis for $\Span(S)$ where $S=
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
,
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
\right\}$.

Solution.

We will first use the leading $1$ method. Consider the system
\[
x_{1}
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
+x_{2}
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
+x_{3}
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
+x_{4}
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
.\] The augmented matrix for this system is
\[
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
2 & -2 & 6 & 1 & 0 \\
1 & -1 & -2 & 3 & 0
\end{array}\right] \xrightarrow{\substack{R_{2}-2R_{1} \\ R_{3}-R_{1}}}
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
0 & 0 & 2 & -1 & 0 \\
0 & 0 & -4 & 2 & 0
\end{array}\right] \] \[
\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 0 \\
0 & 0 & 1 & -1/2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right].
\] Since the above matrix has leading $1$’s in the first and third columns, we can conclude that the first and third vectors of $S$ form a basis of $\Span(S)$. Thus
\[
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
\right\}
\] is a basis for $\Span(S)$.

The post Find a basis for $\Span(S)$, where $S$ is a Set of Four Vectors first appeared on Problems in Mathematics.

For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$
\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]

For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by
\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]

(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.

(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as
\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]

(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

Proof.

(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

We will prove that each vector space axioms holds for $\Fun(S, V)$.

Closed under addition and scalar multiplication: For $f , g \in \Fun(S, V)$ and $c, d \in \K$, $cf+dg$ is another function with domain $S$ defined by
\[ (cf+dg)(s) = cf(s) + dg(s) \in V \, \mbox{ for all } s \in S . \] Thus $cf+dg \in \Fun(S, V)$. This proves that $\Fun(S, V)$ is closed under addition and scalar multiplication.

Addition is associative: For $f, g, h \in \Fun(S, V)$ and $s \in S$ we have
\begin{align*}
((f+g)+h)(s) &= (f+g)(s) + h(s) \\
&= ( f(s) + g(s) ) + h(s) \\
&= f(s) + ( g(s) + h(s) ) \\
&= f(s) + (g+h)(s) \\
&= (f+(g+h))(s) . \end{align*}

Because $((f+g)+h)(s) = (f+(g+h))(s)$ holds for each $s \in S$, we can say that $(f+g)+h = f+(g+h)$.

Addition is commutative: Notice that for $s \in S$ we have
\[ (f+g)(s) = f(s) + g(s) = g(s) + f(s) = (g+f)(s) . \] Because this holds for every $s \in S$, we can say that $f+g = g+f$.

Zero vector: The zero vector is the function $\mathbf{0}$, defined by $\mathbf{0}(s) = 0 \in V$ for each $s \in S$. To check that $\mathbf{0}$ is the additive identity, we check for any $f \in \Fun(S, V)$ and $s \in S$,
\[ (f+ \mathbf{0})(s) = f(s) + \mathbf{0}(s) = f(s) + 0 = f(s) . \] Because this holds for each $s \in S$, we can say that $f + \mathbf{0} = f$.

Inverse vectors: For $f \in \Fun(S, V)$, its additive inverse is the function $(-f)$ defined by
\[ (-f)(s) = – f(s) \, \mbox{ for all } s \in S . \] To check that $-f$ is the inverse of $f$, we check for all $s \in S$
\[ (f + (-f))(s) = f(s) + (-f)(s) = f(s) – f(s) = 0 = \mathbf{0}(s) . \]

Scalar multiplication is associative: For $c, d \in \K$, $f \in \Fun(S, V)$ and $s \in S$ we have
\[ ((cd)(f))(s) = (cd) f(s) = c ( d f(s) ) = c ( df)(s) = (c (df) )(s) . \] Because this holds for each $s \in S$, we conclude that $(cd)(f) = c (df)$.

Scalar identity element: Let $1 \in \K$ be the multiplicative identity. Then for $f \in \Fun(S, V)$ and $s \in S$ we have
\[ (1f)(s) = 1 f(s) = f(s) . \] Because this holds for all $s \in S$, we conclude that $1f = f$.

Distributivity: Next we check distributivity. For $c, d \in \K$, $f \in \Fun(S, V)$ and $s \in S$, we have
\[ ( (c+d)(f) )(s) = (c+d) f(s) = c f(s) + d f(s) = (cf + df)(s) . \] Because this holds for all $s \in S$, we can conclude that
\[ (c+d)f = cf + df . \]

For $c \in \K$, $f, g \in \Fun(S, V)$ and $s \in S$ we have
\begin{align*}
( c(f+g))(s) &= c ( (f+g)(s) ) \\
&= c ( f(s) + g(s) ) \\
&= cf(s) + cg(s) \\
&= (cf + cg)(s) . \end{align*}

Because this holds for all $s \in S$, we conclude that
\[ c(f+g) = cf + cg . \]

We have shown that $\Fun(S, V)$ satisfies all of the vector space axioms, so it is a vector space.

(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

Define the map $T : \Fun(S_1 , V) \rightarrow V$ by
\[ T(f) = f(s) \, \mbox{ for all } f \in \Fun(S_1 , V) . \]

We check that $T$ is a linear map. For $f, g \in \Fun(S_1, V)$ we have
\[ T(f+g) = (f+g)(s) = f(s) + g(s) = T(f) + T(g) . \] Thus $T$ is additive. Next, for $c \in \K$ and $f \in \Fun(S_1, V)$ we have
\[ T(cf) = (cf)(s) = c f(s) = c T(f) . \] This shows that $T$ respects scalar multiplication, and so $T$ is a linear map.

Next we prove that $T$ is an isomorphism. To see that $T$ is surjective, pick any $v \in V$. Define the map $f_v \in \Fun(S_1, V)$ by $f_v(s) = v$. Then $T(f_v) = f_v(s) = v$, and so $v$ lies in the range of $T$.

To see that $T$ is injective, it is enough to show that if $T(f) = 0 \in V$ then $f = \mathbf{0} \in \Fun(S_1, V)$. So suppose that $T(f) = 0$. Because $T(f) = f(s)$, this implies that $f(s) = 0$. But $\mathbf{0}(s) = 0$ as well. Because $f(s) = \mathbf{0}(s)$ for each $s \in S_1$ (which, remember, contains only the one element), we can conclude that $f = \mathbf{0}$.

We have shown that $T$ is linear, injective, and surjective. This finishes the proof that $T$ is a linear isomorphism.

(c) Find a basis of $\Fun(S_1 , V)$.

We can use the basis $B$ of $V$ and the linear isomorphism $T$ found in the previous part to define a basis of $\Fun(S_1, V)$. More specifically, the image of $B$ under the inverse isomorphism
\[ T^{-1} : V \rightarrow \Fun(S_1 , V) \] will define a basis of $\Fun(S_1 , V)$.

For $v \in V$ define the function $f_v$ by $f_v(s) = v$. It is clear that $T(f_v) = v$, and so $T^{-1}(v) = f_v$ for all $v \in V$. In particular, $T^{-1}(e_i) = f_{e_i}$ for each $1 \leq i \leq n$, and so the set $\{ f_{e_1} , f_{e_2} , \cdots , f_{e_n} \}$ forms a basis of $\Fun(S, V)$.

(d) Construct a linear isomorphism between $\Fun(S, V)$ and the vector space $V^m$

Define the map $T : \Fun(S, V) \rightarrow V^m$ by
\[ T(f) = ( f(s_1) , f(s_2) , \cdots , f(s_m) ) \in V^m \, \mbox{ for all } f \in \Fun(S, V) . \]

First we show that $T$ is linear. For $f, g \in \Fun(S, V)$ and $c, d \in \K$ we have
\begin{align*}
T(cf+dg) &= \left( (cf + dg)(s_1) , (cf + dg)(s_2) , \cdots , (cf + dg)(s_m) \right) \\
&= \left( (cf)(s_1) + (dg)(s_1) , (cf)(s_2) + (dg)(s_2) , \cdots , (cf)(s_m) + (dg)(s_m) \right) \\
&= \left( c f(s_1) , c f(s_2) , \cdots , c f(s_m) \right) + \left( d g(s_1) , d g(s_2) , \cdots , d g(s_m) \right) \\
&= c \left( f(s_1) , f(s_2) , \cdots , f(s_m) \right) + d \left( g(s_1) , g(s_2) , \cdots , g(s_m) \right) \\
&= c T(f) + d T(g) . \end{align*}

This proves that $T$ is a linear map.

Next we show that it is an isomorphism.

First we show surjectivity. For $(v_1 , v_2 , \cdots , v_m) \in V^m$ define the function $f$ by
\[ f(s_i) = v_i \, \mbox{ for } 1 \leq i \leq m . \]

Then
\[ T(f) = ( f(s_1) , f(s_2) , \cdots , f(s_m) ) = ( v_1 , v_2 , \cdots , v_m ) . \] Thus every element of $V^m$ lies in the range of $T$.

Next we check injectivity. Suppose that $T(f) = (0 , 0 , \cdots , 0) \in V^m$. This means that $f(s_i) = 0$ for each $1 \leq i \leq m$. Thus $f(s) = \mathbf{0}(s)$ for all $s \in S$, and so $f = \mathbf{0}$. This proves that $T$ is injective. Together with surjectivity and linearity, we have proven that $T$ is an isomorphism.

(e) Constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

We will use the basis $B = \{ e_1 , \cdots , e_n \}$ of $V$ to create a basis for $V^m$. We will translate this basis into a basis of $\Fun(S, V)$ using the isomorphism $T$, or actually its inverse $T^{-1}$.

For integers $i, j$ with $1 \leq i \leq n$ and $1 \leq j \leq m$, define the vector
\[ d_i^j = ( 0 , \cdots , e_i , \cdots , 0 ) . \] Notice that every component is $0$ except for the $j$-th component, which is the vector $e_i$. The set $\{ d_i^j \}_{ \substack{1 \leq i \leq n \\ 1 \leq j \leq m} }$ is the standard basis of $V^m$ defined by $B$. The set $\{ T^{-1} ( d_i^j ) \}_{ \substack{ 1 \leq i \leq n \\ 1 \leq j \leq m}}$ will then be a basis of $\Fun(S, V)$.

To find $T^{-1}$, suppose that $T(f) = d_i^j$. Because the $j$-th component of $d_i^j$ is $e_i$, we must have that $f(s_j) = e_i$. Because all of the other components are $0$, we must have $f(s_k) = 0$ for all $k \neq j$. This information completely determines the function $f$. Define the function $\delta_i^j$ by
\[ \delta_i^j ( s_k ) = \left\{ \begin{array}{cl} e_i & \mbox{ if } k = j \\ 0 & \mbox{ if } k \neq j \end{array} \right. . \]

Then it is clear that $T(\delta_i^j) = d_i^j$; that is, $T^{-1}(d_i^j) = \delta_i^j$. Thus a basis of $\Fun(S, V)$ is formed by the set $\{ \delta_i^j \}_{ \substack{ 1 \leq i \leq n \\ 1 \leq j \leq m }}$. There are exactly $nm$ elements in this basis, thus $\Fun(S, V)$ has dimension $nm$.

(f) Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

We will show in one step that $\Fun(S, W)$ is closed under addition and scalar multiplication. For $f, g \in \Fun(S, W)$, $c, d \in \K$ and $s \in S$ we have
\[ (cf + dg)(s) = c f(s) + d g(s) \in W . \] This proves that $(cf + dg) : S \rightarrow W$, and so $(cf + dg) \in \Fun(S, W)$.

The post Vector Space of Functions from a Set to a Vector Space first appeared on Problems in Mathematics.

Let $A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.

(b) Find a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Solution.

We first obtain the reduced row echelon form matrix corresponding to the matrix $A$.
We reduce the matrix $A$ as follows:
\begin{align*}
A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}\\[6pt] \xrightarrow[R_3-R_1]{R_2-R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
0 &1 & -3 & 1 \\
0 & -1 & 3 & -1
\end{bmatrix}
\xrightarrow[R_3+R_2]{R_1-2R_2}
\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form. That is,
\[\rref(A)=\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}. \tag{*}\]

(a) Find a basis for the nullspace of $A$.

By the computation above, we see that the general solution of $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
x_1&=-9x_3-2x_4\\
x_2&=3x_3-x_4,
\end{align*}
where $x_3$ and $x_4$ are free variables.
Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=\begin{bmatrix}
-9x_3-2x_4 \\
3x_3-x_4 \\
x_3 \\
x_4
\end{bmatrix}
=x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}.
\end{align*}
It follows that the nullspace of the matrix $A$ is given by
\begin{align*}
\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{x}= x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^4 \right \}\\[6pt] &= \Span \left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus, the set
\[\left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}\] is a spanning set for the nullspace $\calN(A)$.
It is straightforward to see that this set is linearly independent, and hence it is a basis for $\calN(A)$.

(b) Find a basis for the row space of $A$.

Recall that the nonzero rows of $\rref(A)$ form a basis for the row space of $A$. (The row space method.)

Thus,
\[\left\{\begin{bmatrix}
1 \\
0 \\
9 \\
2
\end{bmatrix}, \quad \begin{bmatrix}
0 \\
1 \\
-3 \\
1
\end{bmatrix} \right \}\] is a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors of $A$.

Recall that by the leading 1 method, the columns of $A$ corresponding to columns of $\rref(A)$ that contain leading 1 entries form a basis for the range $\calR(A)$ of $A$.
From (*), we see that the first and the second columns contain the leading 1 entries. Thus,
\[\left\{\begin{bmatrix}
2 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
4 \\
3 \\
1
\end{bmatrix}\right \}\] is a basis for the range $\calR(A)$ of $A$.

(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Let us write $A_1, A_2, A_3$, and $A_4$ for the column vectors of the matrix $A$.
In part (c), we showed that $\{A_1, A_2\}$ is a basis for the range $\calR(A)$.
Thus, we need to express the vectors $A_3$ and $A_4$ as a linear combination of $A_1$ and $A_2$, respectively.

A shortcut is to note that the entries of third column vector of $\rref(A)$ give the coefficients of the linear combination for $A_3$. That is, we have
\[A_3=9A_1-3A_2.\] Similarly, the entries of the fourth column of $\rref(A)$ yield
\[A_4=2A_1+A_2.\] Click here if solved 94

The post Find a Basis for Nullspace, Row Space, and Range of a Matrix first appeared on Problems in Mathematics.

Using the definition of the range of a matrix, describe the range of the matrix
\[A=\begin{bmatrix}
2 & 4 & 1 & -5 \\
1 &2 & 1 & -2 \\
1 & 2 & 0 & -3
\end{bmatrix}.\]

Solution.

By definition, the range $\calR(A)$ of the matrix $A$ is given by
\[\calR(A)=\left \{ \mathbf{b} \in \R^3 \quad \middle | \quad A\mathbf{x}=\mathbf{b} \text{ for some } \mathbf{x} \in \R^4 \right \}.\]

Thus, a vector $\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}$ in $\R^3$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.
So, let us find the conditions on $\mathbf{b}$ so that the system is consistent.

To do this, we consider the augmented matrix of the system and reduce it as follows.
\begin{align*}
\left[\begin{array}{rrrr|r}
2 & 4 & 1 & -5 & b_1\\
1 &2 & 1 & -2 & b_2 \\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
2 & 4 & 1 & -5 & b_1\\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow[R_3-R_1]{R_2-2R_1}\\[6pt] \left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
0 & 0 & -1 & -1 & b_1-2b_2\\
0 & 0 & 0 & -1 & b_3-b_2
\end{array}\right] \xrightarrow{-R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
0 & 0 & 1 & 1 & -b_1+2b_2\\
0 & 0 & -1 & -1 & b_3-b_2
\end{array}\right]\\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 0 & -3 & b_1-b_2 \\
0 & 0 & 1 & 1 & -b_1+2b_2\\
0 & 0 & 0 & 0 & -b_1 +b_2 +b_3
\end{array}\right].
\end{align*}

Note that if the $(3, 5)$-entry $-b_1+b_2+b_3$ is not zero, then the system $A\mathbf{x}=\mathbf{0}$ is inconsistent because this implies $0=1$.
On the other hand, if $-b_1+b_2+b_3=0$, then we see that the system is consistent.
Hence, the vector $\mathbf{b}$ is in the range $\calR(A)$ if and only if $-b_1+b_2+b_3=0$.

In summary, we have
\[\calR(A)=\left\{ \begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}\in \R^3 \quad \middle | \quad -b_1+b_2+b_3=0 \right \}.\]

Spanning set for the range

With a little bit additional computation, we can find the spanning set for the range as follows.

Thus, $\mathbf{b} \in \calR(A)$ if and only if
\begin{align*}
\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}=\begin{bmatrix}
b_2+b_3 \\
b_2 \\
b_3
\end{bmatrix}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}.
\end{align*}

In summary, we have
\begin{align*}
\calR(A)&=\left\{ \mathbf{b} \in \R^3 \quad \middle | \quad \mathbf{b}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}\\[6pt] &=\Span\left\{ \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Hence, the spanning set is
\[ \left\{ \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}.\]

This spanning set is linearly independent, hence it’s a basis for the range.

Thus, the dimension of the range is $2$.

The post Describe the Range of the Matrix Using the Definition of the Range first appeared on Problems in Mathematics.

Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis
\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\] Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.

Solution.

We will use the standard basis $C = \{ 1 , x , x^2 , x^3 \}$ to write the coordinate vectors for $f$ and each element of $B$. We then use the simpler coordinate vectors to find a solution to the equation
\begin{equation}\tag{*}
c_1 ( 1 + x ) + c_2 ( 1 + x^2 ) + c_3 ( x – x^2 + 2 x^3 ) + c_4 ( 1 – x – x^2 ) = -3 + 2 x^3 . \end{equation}

The coordinate vectors are found by writing an element as a linear sum of elements of $C$. The coefficients in the linear sum become the entries in the coordinate vector. Thus we have the following coordinate vectors:
\[[ 1 + x ]_{C} = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} , \, [ 1 + x^2 ]_{C} = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} , \, [ x – x^2 + 2 x^3 ]_{C} = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix},\] \[[ 1 – x – x^2 ]_{C} = \begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix} , \, [ -3 + 2 x^3 ]_{C} = \begin{bmatrix} -3 \\ 0 \\ 0 \\ 2 \end{bmatrix}.\]

Plugging these coordinate vectors into Equation (*), we get the equation
\[c_1 \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ 1 \\ -1 \\ 2 \end{bmatrix} + c_4 \begin{bmatrix} 1 \\ -1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 0 \\ 0 \\ 2 \end{bmatrix}.\]

We can now create the augmented matrix which represents this equation:
\[ \left[\begin{array}{rrrr|r}
1 & 1 & 0 & 1 & -3 \\ 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2
\end{array} \right].\]

Now the equation is solved by reducing the augmented matrix:
\begin{align*}
\left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & -3 \\ 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \xrightarrow{ R_2 – R_1 } \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & -3 \\ 0 & -1 & 1 & -2 & 3 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \xrightarrow{ (-1) R_2 } \left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 & -3 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \\[6pt] \xrightarrow[R_3 – R_2]{ R_1 – R_2 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 0 & 0 & -3 & 3 \\ 0 & 0 & 2 & 0 & 2 \end{array} \right] \xrightarrow[ \frac{1}{2} R_4]{ \left( \frac{-1}{3} \right) R_3 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 & 1 \end{array} \right] \\[6pt] \xrightarrow{ R_3 \leftrightarrow R_4 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 2 & -3 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{array} \right] \xrightarrow[ R_2 – 2 R_4 ]{ R_1 + R_4 } \left[\begin{array}{rrrr|r} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{array} \right] \\[6pt] \xrightarrow[R_2 + R_3]{ R_1 – R_3 } \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & -2 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \end{array} \right] .
\end{align*}

We can now read off the solution $c_1 = -2$, $c_2 = 0$ , $c_3 = 1$, and $c_4 = -1$. We can now plug these coefficients into Equation (*), yielding the equation
\[-2 (1 + x) + ( x – x^2 + 2x^3 ) – ( 1 – x – x^2 ) = -3 + 2x^3.\]

Finally, we use these coefficients to write the coordinate vector for $f$ in terms of the basis $B$:
\[ [ -3 + 2x^3 ]_{B} = \begin{bmatrix} -2 \\ 0 \\ 1 \\ -1 \end{bmatrix}.\] Click here if solved 39

The post The Coordinate Vector for a Polynomial with respect to the Given Basis first appeared on Problems in Mathematics.

Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]

Find a basis for the range of $T$.

Solution.

For any matrix $M \in V$ we can write $T(M)$ as a sum
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = a \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & -3 \end{bmatrix} + b \begin{bmatrix} 1 & 0 \\ 2 & 0 \\ 2 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 0 & -3 \\ -1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 2 \\ -1 & 0 \\ 0 & 0 \end{bmatrix}.\]

From this, we see that any element in the range of $T$ can be written as a linear sum of four elements
\[\mathbf{v}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & -3 \end{bmatrix} , \quad \mathbf{v}_2 = \begin{bmatrix} 1 & 0 \\ 2 & 0 \\ 2 & 0 \end{bmatrix},\] \[\mathbf{v}_3 = \begin{bmatrix} 0 & 0 \\ 0 & -3 \\ -1 & 0 \end{bmatrix} , \quad \mathbf{v}_4 = \begin{bmatrix} 0 & 2 \\ -1 & 0 \\ 0 & 0 \end{bmatrix}.\]

This means that the set $\{ \mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3 , \mathbf{v}_4 \}$ is a basis of $W$ as long as the four vectors are linearly independent. To check this, we will show that the coordinate vectors for the $\mathbf{v}_i$, relative to the standard basis, are linearly independent. The standard basis is composed of the matrices
\[\mathbf{e}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \end{bmatrix},\] \[\mathbf{e}_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_5 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{bmatrix} , \, \mathbf{e}_6 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{bmatrix}.\]

Relative to the standard basis $B = \{ \mathbf{e}_1 , \mathbf{e}_2 , \mathbf{e}_3 , \mathbf{e}_4 , \mathbf{e}_5 , \mathbf{e}_6 \}$, the coordinate vectors for the $\mathbf{v}_i$ are
$$ [ \mathbf{v}_1 ]_{B} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ -3 \end{bmatrix} , \, [ \mathbf{v}_2 ]_{B} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 2 \\ 0 \end{bmatrix} , \, [ \mathbf{v}_3 ]_{B} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ -3 \\ -1 \\ 0 \end{bmatrix} , \, [ \mathbf{v}_4 ]_{B} = \begin{bmatrix} 0 \\ 2 \\ -1 \\ 0 \\ 0 \\ 0 \end{bmatrix} . $$

To show that these are linearly independent, we put these vectors in order and obtain the matrix
\[ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & -3 & 0 \\ 0 & 2 & -1 & 0 \\ -3 & 0 & 0 & 0 \end{bmatrix} . \]

To show linear independence of the columns, it suffices to show that this matrix has rank $4$. To do this, we will row-reduce it.

\begin{align*}
\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & -3 & 0 \\ 0 & 2 & -1 & 0 \\ -3 & 0 & 0 & 0 \end{bmatrix} \xrightarrow[\frac{-1}{3} R_6]{ \substack{ \frac{1}{2} R_2 \\[4pt] \frac{-1}{3} R_4 } } \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{ R_1 \leftrightarrow R_6 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix} \\[6pt] \xrightarrow{ R_6 – R_1 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} \xrightarrow[R_5 – 2 R_6 + R_4]{ R_3 – 2 R_6 + R_2} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} .
\end{align*}

The four columns of this matrix are clearly linearly independent, and so it has rank $4$. Thus the original matrix has rank $4$ as well, and so the set $\{ \mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3 , \mathbf{v}_4 \}$ is linearly independent and is a basis of $W$.

The post Find a Basis for the Range of a Linear Transformation of Vector Spaces of Matrices first appeared on Problems in Mathematics.

For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.

Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,
\[T (f) (x) = x f(x).\]

Prove that $T$ is a linear transformation, and find its range and nullspace.

Solution.

$T$ is a linear transformation

We must show that for polynomials $f, g \in \mathrm{P}_n$ and scalar $c \in \mathbb{R}$, the function $T$ satisfies $T(f+g) = T(f) + T(g)$ and $T(cf) = c T(f)$. The first is checked using associativity of multiplication:
\[T(f+g)(x) = x (f+g)(x) = x f(x) + x g(x) = T(f)(x) + T(g)(x) . \] Similarly,
\[T(cf)(x) = x (cf)(x) = c x f(x) = c T(f)(x).\]

The nullspace of $T$

The nullspace of $T$ is the set of polynomials $f(x)$ such that $T(f) = 0$. That is, $x f(x) = 0$. But this product can only be $0$ if one of the terms being multiplied is $0$. Because $x \neq 0$, we must have that $f(x) = 0$. Thus the nullspace is the trivial vector subspace $\{ 0 \}$.

The range of $T$

The range of $T$ is the set of polynomials of the form $x f(x)$ for an arbitrary polynomial $f \in \mathrm{P}_n$. We can calculate this by calculating $T$ for a basis of $\mathrm{P}_n$. Let $B = \{ 1 , x , x^2 , \cdots , x^n \}$ be a basis of $\mathrm{P}_n$.

We see that $T(x^i) = x^{i+1}$ for $ 0 \leq i \leq n$, and so the image of the basis is the set
\[T( B ) = \{ x , x^2 , x^3 , \cdots , x^{n+1} \}.\] Then the range of $T$ must be the span of this set. Specifically,
\[\mathcal{R} ( T ) = \Span ( T( B ) ) = \Span ( x , x^2 , x^3 , \cdots , x^{n+1} ).\] Click here if solved 37

The post The Range and Nullspace of the Linear Transformation $T (f) (x) = x f(x)$ first appeared on Problems in Mathematics.