Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \text{ and } \begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}.\] Then find the rank of the matrix $A$.

(Purdue University, Linear Algebra Final Exam Problem)
 

Solution.

We first determine the nullity of $A$ and deduce the rank of $A$ by the rank-nullity theorem.
The null space $\calN(A)$ of the matrix $A$ is spanned by
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \text{ and } \begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}.\]

Let us find a basis of the null space $\calN(A)$ among these vectors.
We use the “leading 1 method”.
Form the matrix whose column vectors are these three vectors and we apply elementary row operations as follows.
\begin{align*}
\begin{bmatrix}
1 & 2 & 1 \\
2 &1 &-1 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & 2 & 1 \\
0 &-3 & -3 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{-\frac{1}{3}R_2}\\[10pt] \begin{bmatrix}
1 & 2 & 1 \\
0 &1 & 1 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_1-2R_2}
\begin{bmatrix}
1 & 0 & -1 \\
0 &1 & 1 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and the first and second column contain the leading 1’s.

Therefore, the first two vectors
\[\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}\] form a basis of the null space $\calN(A)$.
Hence the nullity, which is the dimension of $\calN(A)$, is $2$.

Since the size of the matrix $A$ is $7 \times 3$, the rank-nullity theorem gives
\[3=\text{nullity of } A + \text{rank of } A.\] Thus, the rank of $A$ is $1$.

The post Given a Spanning Set of the Null Space of a Matrix, Find the Rank first appeared on Problems in Mathematics.

(Purdue University Linear Algebra Exam)

Which of the following statements are true?

(a) A linear system of four equations in three unknowns is always inconsistent.

(b) A linear system with fewer equations than unknowns must have infinitely many solutions.

(c) If the system $A\mathbf{x}=\mathbf{b}$ has a unique solution, then $A$ must be a square matrix.

Solution.

All of them are false as we explain below.

(a) True or False: A linear system of four equations in three unknowns is always inconsistent.

Consider any homogeneous system of four linear equations and three unknowns. Since a homogeneous system always has the solution $\mathbf{x}=\mathbf{0}$. Thus the statement (a) is false.

As an explicit example, the homogeneous system
\[\left\{
\begin{array}{c}
x+y+z=0 \\
2x+2y+2z=0 \\
3x+3y+3z=0
\end{array}
\right.
\] has the solution $(x,y,z)=(0,0,0)$. So the system is consistent.

(b) True or False: A linear system with fewer equations than unknowns must have infinitely many solutions.

Consider the system of one equation with two unknowns
\[0x+0y=1.\] This system has no solution at all. Hence the statement is false.

(c) True or False:If the system $A\mathbf{x}=\mathbf{b}$ has a unique solution, then $A$ must be a square matrix.

Consider the matrix $A=\begin{bmatrix}
1 \\
1
\end{bmatrix}$. Then the system
\[\begin{bmatrix}
1 \\
1
\end{bmatrix}[x]=\begin{bmatrix}
0 \\
0
\end{bmatrix}\] has the unique solution $x=0$ but $A$ is not a square matrix.

A more theoretical argument is as follows. If vectors $\mathbf{v}_1,\dots, \mathbf{v}_k$ are linearly independent, then the system
\[[\mathbf{v}_1 \dots \mathbf{v}_l]\mathbf{x}=\mathbf{0}\] has the unique solution

The post True or False Quiz About a System of Linear Equations first appeared on Problems in Mathematics.

If $L:\R^2 \to \R^3$ is a linear transformation such that
\begin{align*}
L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix}\right)
=\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, \,\,\,\,
L\left( \begin{bmatrix}
1 \\
1
\end{bmatrix}\right)
=\begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}.
\end{align*}
then

(a) find $L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)$, and

(b) find the formula for $L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)$.

If you think you can solve (b), then skip (a) and solve (b) first and use the result of (b) to answer (a).

(Part (a) is an exam problem of Purdue University)

Hint.

1. Express $\begin{bmatrix}
   1 \\
   2
   \end{bmatrix}$ as a linear combination of $\begin{bmatrix}
   1 \\
   0
   \end{bmatrix}$ and $\begin{bmatrix}
   1 \\
   1
   \end{bmatrix}$.
2. Use the linearity of linear transformation $L$.
3. Same for part (b). Replace $\begin{bmatrix}
   1 \\
   2
   \end{bmatrix}$ with the general vector $\begin{bmatrix}
   x \\
   y
   \end{bmatrix}$.

Solution.

(a) Find $L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)$

Note that the vectors $\begin{bmatrix}
1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
1 \\
1
\end{bmatrix}$ are a basis of $\R^2$.
We first express $\begin{bmatrix}
1 \\
2
\end{bmatrix}$ as a linear combination of the vectors $\begin{bmatrix}
1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
1 \\
1
\end{bmatrix}$.
Let $\begin{bmatrix}
1 \\
2
\end{bmatrix}
=c_1 \begin{bmatrix}
1 \\
0
\end{bmatrix} +
c_2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
c_1+c_2 \\
c_2
\end{bmatrix}$.
Solving this, we have $c_1=-1$ and $c_2=2$.

Then we calculate
\begin{align*}
L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)
& =L\left( – \begin{bmatrix}
1 \\
0
\end{bmatrix} +
2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
\right)
= -L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix} \right) +
2 L \left(\begin{bmatrix}
1 \\
1
\end{bmatrix}
\right) \\
&= -\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}
+2 \begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}
=\begin{bmatrix}
3 \\
5 \\
2
\end{bmatrix},
\end{align*}
where the second equality follows from the linearity of $L$.

Thus we have
\[L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)=\begin{bmatrix}
3 \\
5 \\
2
\end{bmatrix}.\]

(b) Find the formula for $L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)$.

 We generalize the proof of (a).
Let $\begin{bmatrix}
x \\
y
\end{bmatrix}
=c_1 \begin{bmatrix}
1 \\
0
\end{bmatrix} +
c_2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
c_1+c_2 \\
c_2
\end{bmatrix}$ be a linear combination. Solving this, we have $c_1=x-y, c_2=y$.

Hence the linear combination is
\[ \begin{bmatrix}
x \\
y
\end{bmatrix}
=(x-y) \begin{bmatrix}
1 \\
0
\end{bmatrix} +
y \begin{bmatrix}
1 \\
1
\end{bmatrix}.
\] Using the linearity of $L$, we compute
\begin{align*}
L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)
&=
L\left( (x-y) \begin{bmatrix}
1 \\
0
\end{bmatrix} +
y \begin{bmatrix}
1 \\
1
\end{bmatrix}\right)
=(x-y) L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix} \right) +
y L\left( \begin{bmatrix}
1 \\
1
\end{bmatrix}\right) \\
&=(x-y) \begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}+y \begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}
=
\begin{bmatrix}
x+y \\
x+2y \\
2x
\end{bmatrix}.
\end{align*}
Therefore the formula is
\[ L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)
= \begin{bmatrix}
x+y \\
x+2y \\
2x
\end{bmatrix}.\] Click here if solved 45

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