# 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. Add to solve later

## 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. Add to solve later

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