basis | Problems in Mathematics

Determine Bases for Nullspaces $\calN(A)$ and $\calN(A^{T}A)$

Yu — Thu, 08 Mar 2018 04:29:35 +0000

Problem 713

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 the Subspace spanned by Five Vectors

Yu — Mon, 26 Feb 2018 13:54:44 +0000

Problem 709

Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[
\mathbf{v}_{1}=
\begin{bmatrix}
1 \\ 2 \\ 2 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{2}=
\begin{bmatrix}
1 \\ 3 \\ 1 \\ 1
\end{bmatrix}
,\;\mathbf{v}_{3}=
\begin{bmatrix}
1 \\ 5 \\ -1 \\ 5
\end{bmatrix}
,\;\mathbf{v}_{4}=
\begin{bmatrix}
1 \\ 1 \\ 4 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{5}=
\begin{bmatrix}
2 \\ 7 \\ 0 \\ 2
\end{bmatrix}
.\] Find a basis for the span $\Span(S)$.

We will give two solutions.

Solution 1.

We apply the leading 1 method.
Let $A$ be the matrix whose column vectors are vectors in the set $S$:
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Applying the elementary row operations to $A$, we obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
\xrightarrow[R_4+R_1]{\substack{R_2-2R_1 \\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
0 & 1 & 3 & -1 & 3 \\
0 & -1 & -3 & 2 & -4 \\
0 & 2 & 6 & 0 & 4
\end{bmatrix}\\[6pt] \xrightarrow[R_4-2R_2]{\substack{R_1-R_2 \\ R_3+R_2}}
\begin{bmatrix}
1 & 0 & -2 & 2 & -1 \\
0 & 1 & 3 & -1 & 3 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 2 & -2
\end{bmatrix}
\xrightarrow[R_4-2R_3]{\substack{R_1-2R_3 \\ R_2+R_3}}
\begin{bmatrix}
1 & 0 & -2 & 0 & 1 \\
0 & 1 & 3 & 0 & 2 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}=\rref(A).
\end{align*}
Observe that the first, second, and fourth column vectors of $\rref(A)$ contain the leading 1 entries.
Hence, the first, second, and fourth column vectors of $A$ form a basis of $\Span(S)$.
Namely,
\[\left\{ \begin{bmatrix}
1 \\
2 \\
2 \\
-1
\end{bmatrix}, \begin{bmatrix}
1 \\
3 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
4 \\
-1
\end{bmatrix}\right \}\] is a basis for $\Span(S)$.

Solution 2.

Let
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Then $\Span(S)$ is the column space of $A$, which is the row space of $A^{T}$. Using row operations, we have
\[
A^{T}=
\begin{bmatrix}
1 & 2 & 2 & -1 \\
1 & 3 & 1 & 1 \\
1 & 5 & -1 & 5 \\
1 & 1 & 4 & -1 \\
2 & 7 & 0 & 2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & 2 & -1 \\
0 & 1 & -1 & 2 \\
0 & 3 & -3 & 6 \\
0 & -1 & 2 & 0 \\
0 & 3 & -4 & 4
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 4 & 5 \\
0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & -1 & -2
\end{bmatrix}
\] \[
\to
\begin{bmatrix}
1 & 0 & 0 & -13 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
.
\] Therefore, the set of nonzero rows
\[
\left\{
\begin{bmatrix}
1 \\ 0 \\ 0 \\ -13
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 1 \\ 0 \\ 4
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 0 \\ 1 \\ 2
\end{bmatrix}
\right\}
\] is a basis for the row space of $A^{T}$, which equals $\Span(S)$.

The post Find a Basis for the Subspace spanned by Five Vectors first appeared on Problems in Mathematics.

The Coordinate Vector for a Polynomial with respect to the Given Basis

Yu — Mon, 29 Jan 2018 04:39:09 +0000

Problem 683

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 $B$:
\[ [ -3 + 2x^3 ]_{B} = \begin{bmatrix} -2 \\ 0 \\ 1 \\ -1 \end{bmatrix}.\] The post The Coordinate Vector for a Polynomial with respect to the Given Basis first appeared on Problems in Mathematics.

Find the Matrix Representation of $T(f)(x) = f(x^2)$ if it is a Linear Transformation

Yu — Mon, 22 Jan 2018 03:05:09 +0000

Problem 679

For an integer $n > 0$, let $\mathrm{P}_n$ denote the vector space of polynomials with real coefficients of degree $2$ or less. Define the map $T : \mathrm{P}_2 \rightarrow \mathrm{P}_4$ by
\[ T(f)(x) = f(x^2).\]

Determine if $T$ is a linear transformation.

If it is, find the matrix representation for $T$ relative to the basis $\mathcal{B} = \{ 1 , x , x^2 \}$ of $\mathrm{P}_2$ and $\mathcal{C} = \{ 1 , x , x^2 , x^3 , x^4 \}$ of $\mathrm{P}_4$.

Solution.

$T$ is a linear transformation

To prove that $T$ is a linear transformation, we must show that it satisfies both axioms for linear transformations. For $f, g \in \mathrm{P}_2$, we have
\[T( f+g )(x) = (f+g)(x^2) = f(x^2) + g(x^2) = T(f)(x) + T(g)(x)\] while for a scalar $c \in \mathbb{R}$, we have
\[ T( c f )(x) = (cf)(x^2) = c f(x^2) = c T(f)(x).\] We see that $T$ is a linear transformation.

The matrix representation for $T$

To find its matrix representation, we must calculate $T(f)$ for each $f \in \mathcal{B}$ and find its coordinate vector relative to the basis $\mathcal{C}$. We calculate
\[T(1) = 1 , \quad T(x) = x^2 , \quad T(x^2) = x^4.\] Each of these is an element of $\mathcal{C}$. Their coordinate vectors relative to $\mathcal{C}$ are thus
\[[ T(1) ]_{\mathcal{B}} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} , \quad [ T(x) ]_{\mathcal{B}} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} , \quad [ T(x^2) ]_{\mathcal{C}} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}.\]

The matrix representation of $T$ is found by combining these columns, in order, into one matrix:

\[ [T]_{\mathcal{B}}^{\mathcal{C}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\] The post Find the Matrix Representation of $T(f)(x) = f(x^2)$ if it is a Linear Transformation first appeared on Problems in Mathematics.

The Matrix Representation of the Linear Transformation $T (f) (x) = ( x^2 – 2) f(x)$

Yu — Tue, 16 Jan 2018 05:10:05 +0000

Problem 673

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 $T : \mathrm{P}_3 \rightarrow \mathrm{P}_{5}$ be the map defined by, for $f \in \mathrm{P}_3$,
\[T (f) (x) = ( x^2 – 2) f(x).\]

Determine if $T(x)$ is a linear transformation. If it is, find the matrix representation of $T$ relative to the standard basis of $\mathrm{P}_3$ and $\mathrm{P}_{5}$.

Solution.

We must check that $T$ satisfies the two axioms for linear transformations. Suppose $f, g \in \mathrm{P}_3$. Then
\[T(f+g)(x) = (x^2-2) (f+g)(x) = (x^2 – 2) f(x) + (x^2 – 2) g(x) = T(f)(x) + T(g)(x).\]

Second, if $c \in \mathbb{R}$ then
\[T( cf )(x) = (x^2 – 2) ( cf) (x) = c (x^2 – 2) f(x) = c T(f)(x).\]

The two axioms are satisfied, and so $T$ is a linear transformation.

Now we find its matrix representation relative to the standard basis $B = \{ 1 , x , x^2 , x^3 \}$ of $\mathrm{P}_3$ and the standard basis $C = \{ 1 , x , x^2 , x^3 , x^4 , x^5 \} $ . To do this, for every $f \in B$ we calculate the coordinate vector of $T(f)$ relative to the basis $C$.

For example, we find that $T(1) = x^2 – 2$. The coordinate vector for this element, relative to $C$, is
\[ [ x^2 – 2 ]_{C} = \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}.\] Similarly, we find that $T(x) = x^3 – 2x$. The coordinate vector for this element, relative to $C$,
\[ [ x^3 – 2x ]_{C} = \begin{bmatrix} 0 \\ -2 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}.\]

Performing this process for the rest of the elements of $B$, we get
\[ [T(x^2)]_{C} = [ x^4 – 2x^2 ]_{C} = \begin{bmatrix} 0 \\ 0 \\ -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \quad [T(x^3)]_{C} = [ x^5 – 2x^3 ]_{C} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ -2 \\ 0 \\ 1 \end{bmatrix}.\] And now putting these four column vectors in order, we get the matrix representation for $T$:
\[ [T]_{B}^{C} = \begin{bmatrix} -2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.\] The post The Matrix Representation of the Linear Transformation $T (f) (x) = ( x^2 – 2) f(x)$ first appeared on Problems in Mathematics.

The Range and Nullspace of the Linear Transformation $T (f) (x) = x f(x)$

Yu — Tue, 16 Jan 2018 03:48:31 +0000

Problem 672

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,
\[\mathcal{R} ( T ) = \Span ( T( B ) ) = \Span ( x , x^2 , x^3 , \cdots , x^{n+1} ).\] The post The Range and Nullspace of the Linear Transformation $T (f) (x) = x f(x)$ first appeared on Problems in Mathematics.

The Vector $S^{-1}\mathbf{v}$ is the Coordinate Vector of $\mathbf{v}$

Yu — Wed, 20 Dec 2017 03:40:36 +0000

Problem 632

Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.
Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.

Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.

Proof.

We first express the vector $\mathbf{v}$ as a linear combination of the basis vectors
\[\mathbf{v}=c_1\mathbf{v}_+c_2 \mathbf{v}_2.\] This expression is unique and the coordinate vector of $\mathbf{v}$ with respect to the basis $B$ is defined to be
\[[\mathbf{v}]_B =\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}.\]

Let
\[S^{-1}\mathbf{v}= \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}.\] Or equivalently,
\[\mathbf{v}=S\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}.\] Our goal is to show that $\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}$.

We have
\begin{align*}
c_1\mathbf{v}_+c_2 \mathbf{v}_2&=\mathbf{v}=S\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=x_1\mathbf{v}_1+x_2\mathbf{v}_2.
\end{align*}

Hence
\[(x_1-c_1)\mathbf{v}_1+(x_2-c_2)\mathbf{v}_2=\mathbf{0}.\] As $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is linearly independent, we obtain $x_1=c_1$ and $x_2=c_2$.

Therefore,
\[S^{-1}\mathbf{v}=\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=[\mathbf{v}]_B.\] The post The Vector $S^{-1}\mathbf{v}$ is the Coordinate Vector of $\mathbf{v}$ first appeared on Problems in Mathematics.

If Matrices Commute $AB=BA$, then They Share a Common Eigenvector

Yu — Tue, 14 Nov 2017 04:59:06 +0000

Problem 608

Let $A$ and $B$ be $n\times n$ matrices and assume that they commute: $AB=BA$.
Then prove that the matrices $A$ and $B$ share at least one common eigenvector.

Proof.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$. That is, we have $A\mathbf{x}=\lambda \mathbf{x}$.
Then we claim that the vector $\mathbf{v}:=B\mathbf{x}$ belongs to the eigenspace $E_{\lambda}$ of $\lambda$.
In fact, as $AB=BA$ we have
\begin{align*}
A\mathbf{v}=AB\mathbf{x}=BA\mathbf{x} =\lambda B\mathbf{x}=\lambda \mathbf{v}.
\end{align*}
Hence $\mathbf{v}\in E_{\lambda}$.

Now, let $\{\mathbf{x}_1, \dots, \mathbf{x}_k\}$ be an eigenbasis of the eigenspace $E_{\lambda}$.
Set $\mathbf{v}_i=B\mathbf{x}_i$ for $i=1, \dots, k$.
The above claim yields that $\mathbf{v}_i \in E_{\lambda}$, and hence we can write
\[\mathbf{v}_i=c_{1i} \mathbf{x}_1+c_{2i}\mathbf{x}_2+\cdots+c_{ki}\mathbf{x}_k \tag{*}\] for some scalars $c_{1i}, c_{2i}, \dots, c_{ki}$.

Extend the basis $\{\mathbf{x}_1, \dots, \mathbf{x}_k\}$ of $E_{\lambda}$ to a basis
\[\{\mathbf{x}_1, \dots, \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n\}\] of $\R^n$ by adjoining vectors $\mathbf{x}_{k+1}, \dots, \mathbf{x}_n$.

Then we obtain using (*)
\begin{align*}
&B [\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n]\\
&=[B\mathbf{x}_1,\dots , B\mathbf{x}_k, B\mathbf{x}_{k+1}, \dots, B\mathbf{x}_n]\\
&=[\mathbf{v}_1,\dots , \mathbf{v}_k, B\mathbf{x}_{k+1}, \dots, B\mathbf{x}_n]\\[6pt] &=[\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n] \left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right],\tag{**}
\end{align*}

where $C=(c_{ij})$ is the $k\times k$ matrix whose entries are the coefficients $c_{ij}$ of the linear combination (*), $O$ is the $(n-k) \times k$ zero matrix, $D$ is a $k \times (n-k)$ matrix, and $F$ is an $(n-k) \times (n-k)$ matrix.

Let $P=[\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n]$.
As the column vectors of $P$ are linearly independent, $P$ is invertible.

From (**), we obtain
\[P^{-1}BP=\left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right].\] It follows that
\begin{align*}
\det(B-tI)&=\det(P^{-1}BP-tI)=\left|\begin{array}{c|c}
C-tI & D\\
\hline
O & F-tI
\end{array}
\right|=\det(C-tI)\det(F-tI).
\end{align*}

Let $\mu$ be an eigenvalue of the matrix $C$ and let $\mathbf{a}$ be an eigenvector corresponding to $\mu$.
Then as $\det(C-\mu I)=0$, we see that $\det(B-\mu I)=0$ and $\mu$ is an eigenvalue of $B$.

Write
\[\mathbf{a}=\begin{bmatrix}
a_1 \\
a_2 \\
\vdots \\
a_k
\end{bmatrix}\neq \mathbf{0}\] and define a new vector by
\[\mathbf{y}=a_1\mathbf{x}_1+\cdots +a_k \mathbf{x}_k\in E_{\lambda}.\] Then $\mathbf{y}$ is an eigenvector in $E_{\lambda}$ since it is a nonzero (as $\mathbf{a}\neq \mathbf{0}$) linear combination of the basis $E_{\lambda}$.

Multiplying $BP=P\left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right]$ from (**) by the $n$-dimensional vector
\[\begin{bmatrix}
\mathbf{a} \\
0 \\
\vdots\\
0
\end{bmatrix}=\begin{bmatrix}
a_1 \\
\vdots \\
a_k \\
0 \\
\vdots\\
0
\end{bmatrix}\] on the right, we have
\[BP\begin{bmatrix}
\mathbf{a} \\
0 \\
\vdots\\
0
\end{bmatrix}=P\left[\begin{array}{c|c}
C & D\\
\hline
O & F
\end{array}
\right] \begin{bmatrix}
\mathbf{a} \\
0 \\
\vdots\\
0
\end{bmatrix}.\]

The left hand side is equal to
\[B[a_1\mathbf{x}_1+\cdots+a_k \mathbf{x}_k]=B\mathbf{y}.\]

On the other hand, the right hand side is equal to
\begin{align*}
P\begin{bmatrix}
C \mathbf{a}\\
\mathbf{0}
\end{bmatrix}
=P\begin{bmatrix}
\mu \mathbf{a} \\
\mathbf{0}
\end{bmatrix}=\mu P\begin{bmatrix}
\mathbf{a} \\
\mathbf{0}
\end{bmatrix}
=\mu [a_1\mathbf{x}_1+\cdots+a_k \mathbf{x}_k]=\mu \mathbf{y}.
\end{align*}

Therefore, we obtain
\[B\mathbf{y}=\mu \mathbf{y}.\] This proves that the vector $\mathbf{y}$ is eigenvector of both $A$ and $B$.
Hence, this completes the proof.

The post If Matrices Commute $AB=BA$, then They Share a Common Eigenvector first appeared on Problems in Mathematics.

Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less

Yu — Wed, 08 Nov 2017 16:25:29 +0000

Problem 607

Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.
Let
\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where
\begin{align*}
p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\
p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.
\end{align*}

(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.

(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.

(The Ohio State University, Linear Algebra Midterm)

Solution.

(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.

Let $B=\{1, x, x^2, x^3\}$ be the standard basis for $\calP_3$.
With respect to the basis $B$, the coordinate vectors of the given polynomials are
\begin{align*}
\mathbf{v}_1:=[p_1(x)]_B&=\begin{bmatrix}
1 \\
3 \\
2 \\
-1
\end{bmatrix}, &\mathbf{v}_2:=[p_2(x)]_B=\begin{bmatrix}
0 \\
1 \\
0 \\
1
\end{bmatrix}\\[6pt] \mathbf{v}_3:=[p_3(x)]_B&=\begin{bmatrix}
0 \\
1 \\
1 \\
-1
\end{bmatrix}, &\mathbf{v}_4:=[p_4(x)]_B=\begin{bmatrix}
3 \\
8 \\
0 \\
8
\end{bmatrix}.
\end{align*}
Let $T=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}$ be the set of these coordinate vectors.

We find a basis for $\Span(T)$ among the vectors in $T$ by the leading 1 method.
We form the matrix whose column vectors are the vectors in $T$ and apply elementary row operations as follows.

\begin{align*}
\begin{bmatrix}
1 & 0 & 0 & 3 \\
3 &1 & 1 & 8 \\
2 & 0 & 1 & 0 \\
-1 & 1 & -1 & 8
\end{bmatrix}
\xrightarrow{\substack{R_2-3R_1\\R_3-2R_1\\R_4+R_1}}
\begin{bmatrix}
1 & 0 & 0 & 3 \\
0 &1 & 1 & -1 \\
0 & 0 & 1 & -6 \\
0 & 1 & -1 & 11
\end{bmatrix}
\xrightarrow{R_4-R_2}\\[6pt] \begin{bmatrix}
1 & 0 & 0 & 3 \\
0 &1 & 1 & -1 \\
0 & 0 & 1 & -6 \\
0 & 0 & -2 & 12
\end{bmatrix}
\xrightarrow{R_4+2R_2}
\begin{bmatrix}
1 & 0 & 0 & 3 \\
0 &1 & 0 & 5 \\
0 & 0 & 1 & -6 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}

The first three columns of the reduced row echelon form matrix contain the leading 1’s.
Thus, $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a basis for $\Span(T)$.
It follows that $Q:=\{p_1(x), p_2(x), p_3(x)\}$ is a basis for $\Span(S)$.

(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.

Note that $p_4(x)$ is not in the basis $Q$.

The fourth column of the matrix in reduced row echelon form of part (a) gives the coefficients of the linear combination:
\[p_4(x)=3p_1(x)+5p_2(x)-6p_3(x).\]

Thus, the coordinate vector of $p_4(x)$ with respect to the basis $Q$ is
\[[p_4(x)]_Q=\begin{bmatrix}
3 \\
5 \\
-6
\end{bmatrix}.\]

Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

In part (b), some students stopped after obtaining the linear combination $p_4(x)=3p_1(x)+5p_2(x)-6p_3(x)$.
You must read the problem carefully. You are asked to find the coordinate vector of $p_4(x)$ with respect to $Q$.

List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

Vector Space of 2 by 2 Traceless Matrices
Find an Orthonormal Basis of the Given Two Dimensional Vector Space
Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less ←The current problem

The post Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less first appeared on Problems in Mathematics.

Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$

Yu — Wed, 08 Nov 2017 12:13:46 +0000

Problem 605

Let $T:\R^2 \to \R^3$ be a linear transformation such that
\[T\left(\, \begin{bmatrix}
3 \\
2
\end{bmatrix} \,\right)
=\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix} \text{ and }
T\left(\, \begin{bmatrix}
4\\
3
\end{bmatrix} \,\right)
=\begin{bmatrix}
0 \\
-5 \\
1
\end{bmatrix}.\]

(a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$).

(b) Determine the rank and nullity of $T$.

(The Ohio State University, Linear Algebra Midterm)

Solution.

(a) Find the matrix representation of $T$.

The matrix representation $A$ of the linear transformation is given by
\[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)],\] where
\[\mathbf{e}_1=\begin{bmatrix}
1\\ 0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0\\ 1
\end{bmatrix}\] are the standard basis vectors for $\R^2$.

To determine $T(\mathbf{e}_1)$, we first express $\mathbf{e}_1$ as a linear combination of $\begin{bmatrix}
3 \\
2
\end{bmatrix}$ and $\begin{bmatrix}
4 \\
3
\end{bmatrix}$ as follows.
Let
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}=a\begin{bmatrix}
3 \\
2
\end{bmatrix}+b\begin{bmatrix}
4 \\
3
\end{bmatrix}.\] This can be written as
\[\begin{bmatrix}
3 & 4\\
2& 3
\end{bmatrix}\begin{bmatrix}
a \\
b
\end{bmatrix}=\begin{bmatrix}
1 \\
0
\end{bmatrix}.\] The determinant of the coefficient matrix $\begin{bmatrix}
3 & 4\\
2& 3
\end{bmatrix}$ is $3\cdot 3-4\cdot 2=1\neq 0$ and thus it is invertible.
Hence we have
\[\begin{bmatrix}
a \\
b
\end{bmatrix}
=\begin{bmatrix}
3 & 4\\
2& 3
\end{bmatrix}^{-1}\begin{bmatrix}
1 \\
0
\end{bmatrix}
=\begin{bmatrix}
3 & -4\\
-2& 3
\end{bmatrix}\begin{bmatrix}
1 \\
0
\end{bmatrix}=\begin{bmatrix}
3 \\
-2
\end{bmatrix}.\] Hence $a=3$ and $b=-2$.
It yields that
\[\mathbf{e}_1=3\begin{bmatrix}
3 \\
2
\end{bmatrix}-2\begin{bmatrix}
4 \\
3
\end{bmatrix}.\]

It follows that
\begin{align*}
T(\mathbf{e}_1)&=T\left(\,3\begin{bmatrix}
3 \\
2
\end{bmatrix}-2\begin{bmatrix}
4 \\
3
\end{bmatrix} \,\right)\\[6pt] &=3T\left(\, \begin{bmatrix}
3 \\
2
\end{bmatrix}\,\right)-2T\left(\, \begin{bmatrix}
4 \\
3
\end{bmatrix} \,\right) &&\text{by linearity of $T$}\\[6pt] &=3\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}-2\begin{bmatrix}
0 \\
-5 \\
1
\end{bmatrix}=\begin{bmatrix}
3 \\
16 \\
7
\end{bmatrix}.
\end{align*}

Similarly, we compute $T(\mathbf{e}_2)$ as follows.
Let
\[\mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}=c\begin{bmatrix}
3 \\
2
\end{bmatrix}+d\begin{bmatrix}
4 \\
3
\end{bmatrix}.\] This can be written as
\[\begin{bmatrix}
3 & 4\\
2& 3
\end{bmatrix}\begin{bmatrix}
a \\
b
\end{bmatrix}=\begin{bmatrix}
0 \\
1
\end{bmatrix}.\] Hence, we obtain
\[\begin{bmatrix}
c \\
d
\end{bmatrix}=\begin{bmatrix}
3 & 4\\
2& 3
\end{bmatrix}^{-1}\begin{bmatrix}
0 \\
1
\end{bmatrix}=\begin{bmatrix}
3 & -4\\
-2& 3
\end{bmatrix}\begin{bmatrix}
0 \\
1
\end{bmatrix}=\begin{bmatrix}
-4 \\
3
\end{bmatrix},\] and $c=-4, d=3$.

Hence
\[\mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}=-4\begin{bmatrix}
3 \\
2
\end{bmatrix}+3\begin{bmatrix}
4 \\
3
\end{bmatrix}\] and we have
\begin{align*}
T(\mathbf{e}_2)&=T\left(\,-4\begin{bmatrix}
3 \\
2
\end{bmatrix}+3\begin{bmatrix}
4 \\
3
\end{bmatrix} \,\right)\\[6pt] &=-4T\left(\,\begin{bmatrix}
3 \\
2
\end{bmatrix}\,\right)+3T\left(\,\begin{bmatrix}
4 \\
3
\end{bmatrix}\,\right) &&\text{by linearity of $T$}\\[6pt] &=-4\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}+3\begin{bmatrix}
0 \\
-5 \\
1
\end{bmatrix}=\begin{bmatrix}
-4 \\
-23 \\
-9
\end{bmatrix}.
\end{align*}

Therefore the matrix representation $A$ of $T$ is
\[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)]=\begin{bmatrix}
3 & -4 \\
16 & -23 \\
7 &-9
\end{bmatrix}.\]

(b) Determine the rank and nullity of $T$.

Note that the rank and nullity of $T$ are the same as the those of the matrix representation $A$ of $T$.
In part (a), we obtained
\[A=\begin{bmatrix}
3 & -4 \\
16 & -23 \\
7 &-9
\end{bmatrix}.\] Let us first determine the rank of $A$.
We have
\begin{align*}
A=\begin{bmatrix}
3 & -4 \\
16 & -23 \\
7 &-9
\end{bmatrix}
\xrightarrow{R_3-2R_1}
\begin{bmatrix}
3 & -4 \\
16 & -23 \\
1 &-1
\end{bmatrix}
\xrightarrow{R_1\leftrightarrow R_3}
\begin{bmatrix}
1 & -1 \\
16 & -23 \\
3 &-4
\end{bmatrix}\\[6pt] \xrightarrow[R_3-3R_1]{R_2-16R_1}
\begin{bmatrix}
1 & -1 \\
0 & -7 \\
0 &-1
\end{bmatrix}
\xrightarrow{-R_3}
\begin{bmatrix}
1 & -1 \\
0 & -7 \\
0 &1
\end{bmatrix}
\xrightarrow[R_2+7R_3]{R_1+R_3}
\begin{bmatrix}
1 & 0 \\
0 & 0 \\
0 &1
\end{bmatrix}
\xrightarrow{R_2\leftrightarrow R_3}
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 &0
\end{bmatrix}.
\end{align*}
Hence, the reduced row echelon form matrix of $A$ has two nonzero rows.
So the rank of $A$ is $2$.

By the rank-nullity theorem, we know that
\[\text{(rank of $A$)+(nullity of $A$)}=2.\] As the rank of $A$ is $2$, we see that the nullity of $A$ is $0$.

Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

Vector Space of 2 by 2 Traceless Matrices
Find an Orthonormal Basis of the Given Two Dimensional Vector Space
Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$ ←The current problem
Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less

The post Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$ first appeared on Problems in Mathematics.