Given matrix times vector, find the inverse of the matrix times the same vector.

Question

Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.

My thought are the following:

Step 1:

Times $A^{-1}$ for both sides.

$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$

$V=8A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$

$\frac{1}{8} V=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$

Step 2:

Can I assume $AV=\lambda V=8*\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$ here? Hence, $V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}$? Is this a correct step?

Step 3:

If step 2 is correct, then

$\frac{1}{8}V = \frac{1}{8} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} \begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}=A^{-1} V=\begin{bmatrix} \frac{1}{8} \\ \frac{1}{8} \\ \frac{1}{8} \\\frac{1}{8}\end{bmatrix}$

Does the above inference sounds right?

Answer 1

In general, no you can't assume $AV = \lambda V$. Not every vector is an eigenvector. Even if $V$ were an eigenvector, then this would still not give you a well-defined answer.

For one example, take $A = I$, the $4 \times 4$ identity matrix. Then $AV = V$, hence $V = \begin{bmatrix} 8 \\ 8 \\ 8 \\ 8 \end{bmatrix} \neq \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$.

As another example, take $A = 2I$. Then $AV = 2V$, hence $V = \begin{bmatrix} 4 \\ 4 \\ 4 \\ 4 \end{bmatrix} \neq \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$.

As you can see, there clearly needs to be more conditions placed on $A$ before this has a clear solution.

Answer 2

If B=A^-1 and b=<8,8,8,8> then from AV=b you get V=Bb and then BV=(B^2)b. I gess that this is the final result, if there are not other information about A.

Answer 3

What if I add extra information regarding the vector $V$ as the following,

Question: Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and ${V=\begin{bmatrix}1\\1 \\ 1 \\1\end{bmatrix}}$ , find $A^{-1}V$.

Times $A^{-1}$ for both sides.

$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}=A^{-1} * 8 * V$

$V=A^{-1} * 8 * V$

$\frac{1}{8}V = A^{-1} V=\begin{bmatrix} \frac{1}{8} \\ \frac{1}{8} \\ \frac{1}{8} \\\frac{1}{8}\end{bmatrix}$

Add information about $V$ seems become a trivial question?

Explaination:I did not use the assumption that V is the eigenvector of the matrix A here. Just factor out 8 from $[8,8,8,8]^T$ to become $V$.