# Linearly Independent vectors $\mathbf{v}_1, \mathbf{v}_2$ and Linearly Independent Vectors $A\mathbf{v}_1, A\mathbf{v}_2$ for a Nonsingular Matrix

## Problem 284

Let $\mathbf{v}_1$ and $\mathbf{v}_2$ be $2$-dimensional vectors and let $A$ be a $2\times 2$ matrix.

**(a)** Show that if $\mathbf{v}_1, \mathbf{v}_2$ are linearly dependent vectors, then the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are also linearly dependent.

**(b)** If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent vectors, can we conclude that the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are also linearly independent?

**(c)** If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent vectors and $A$ is nonsingular, then show that the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are also linearly independent.

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Contents

- Problem 284
- Proof.
- (a) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly dependent, then $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly dependent.
- (b) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, can we conclude that $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent?
- (c) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent and $A$ is nonsingular, then $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent.

## Proof.

### (a) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly dependent, then $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly dependent.

Since $\mathbf{v}_1, \mathbf{v}_2$ are linearly dependent vectors, there exists $(c_1, c_2)\neq (0,0)$ such that

\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]
Using this equality, we obtain

\begin{align*}

\mathbf{0}&=A\mathbf{0}\\

&=A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)\\

&=A(c_1\mathbf{v}_1)+A(c_2\mathbf{v}_2)\\

&=c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2) \qquad \text{ (since $c_1, c_2$ are scalars)}.

\end{align*}

Hence we have

\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0}\]
with $(c_1, c_2) \neq (0, 0)$.

This implies that the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly dependent.

### (b) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, can we conclude that $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent?

The answer is no. In general, even though $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent vectors, the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ might be linearly dependent. Let us give an example.

Let $\mathbf{v}_1=\begin{bmatrix}

1 \\

0

\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}

0 \\

1

\end{bmatrix}$.

Then it is straightforward to see that these vectors are linearly independent.

Let

\[A=\begin{bmatrix}

0 & 0\\

0& 0

\end{bmatrix}\]
be the $2 \times 2$ zero matrix.

Then we have

\[A\mathbf{v}_1=\begin{bmatrix}

0 \\

0

\end{bmatrix}, A\mathbf{v}_2=\begin{bmatrix}

0 \\

0

\end{bmatrix}\]
and clearly $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly dependent vectors.

### (c) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent and $A$ is nonsingular, then $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent.

Consider the linear combination

\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0}.\]
To show that $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent, we show that $c_1=c_2=0$.

The above equality can be written as

\[A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)=\mathbf{0}.\]
Since the matrix $A$ is nonsingular, the homogeneous equation $A\mathbf{x}=\mathbf{0}$ has only the zero solution. Therefore we have

\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]
(Another reasoning here is that since the matrix $A$ is nonsingular, it is invertible, and thus we have the inverse matrix $A^{-1}$. Multiplying by $A^{-1}$ on the left, we obtain the same equation.)

Now, since $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent by assumption, it follows that $c_1=c_2=0$.

Hence we conclude that the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent.

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