If $Ax = b$ has one solution then $(A^2)x = b$ also has one solution?

Question

This is my start of the proof but I'm stuck:

Given $A$ squared matrix of order $n$ and $B$ is a column of length $n$:

1. Assume that $Ax=B$ has one solution. Let this solution be denoted by $x_0$.

2. Now, we want to show that $A^2x_1 = B$ also has one solution. Let us assume that there exists another solution $x_2$ such that $A^2x_2 = B$.

3. ?

Answer 1

HINT: Suppose that $Ax=b$ has at least one solution i.e., let $x_0$ be such that $Ax_0=b$.

• Then if $A$ is not invertible, the equation $Ax=b$ has more than solution. Indeed, if $A$ is not invertible, then there is a $y \not = 0$ such that the equation $Ay=0$ is satisfied. But then both equations $Ax_0=b$ and $A(x_0+y)=b$ hold, with $x_0 \not = x_0+y$. Make sure you see why. So then if $A$ is not invertible, then the equation $Ax=b$ has more than one solution [assuming the equation $Ax=b$ has at least one solution that is].

• So then $A$ must be invertible for the equation $Ax=b$ to have exactly one solution. But then if $A$ is invertible, then $A^2$ must also be invertible. Now, for any invertible $n \times n$ matrix $B$ and any vector $z$ of length $n$, there is exactly one solution to the equation $Bx=z$....

Can you finish from here.

To revisit your previous question, let $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, and let $b= \begin{bmatrix}1 \\ 0 \end{bmatrix}$. Then the equation $A^2x=b$ has no solutions, but also the equation $Ax=b$ has more than one solution, namely $x= \begin{bmatrix} a \\ 1 \end{bmatrix}$ for any real $a$ will satisfy the equation $Ax=b$.

Answer 2

Your now deleted steps 3 and onward were close.

Step 5 mentioned that, if $A\mathbf x=\mathbf b$ has exactly one solution, then we know that its null space is trivial: $A\mathbf v=\mathbf 0$ also has exactly one solution $\mathbf v =\mathbf 0$.

(Otherwise if $\mathbf v\ne \mathbf 0$, then $A(\mathbf x + \mathbf v) = \mathbf b$, contradicting that $A\mathbf x=\mathbf b$ has exactly one solution.)

Following the attempt in your question, assume $A^2\mathbf x_1=\mathbf b$ and $A^2\mathbf x_2 = \mathbf b$.

3. We know that $A^2 (\mathbf x_2-\mathbf x_1) = A^2\mathbf x_2 - A^2\mathbf x_1 = \mathbf 0$.

4. We know that the null space is trivial. So $A(\mathbf x_2 - \mathbf x_1)$ must be $\mathbf 0$.

5. We know that the null space is trivial. So $(\mathbf x_2 - \mathbf x_1)$ must be $\mathbf 0$.

6. Or $\mathbf x_2 = \mathbf x_1$. This implies that there is only one solution to $A^2\mathbf x = \mathbf b$.