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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?

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    $\begingroup$ If you can find $x/y$ given $xy=8$ (only) then you could solve the stated problem. $\endgroup$ – Math Lover Jan 3 '18 at 4:14
  • $\begingroup$ Do you mean there is some missing statement here? $\endgroup$ – Ying Jan 3 '18 at 4:25
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    $\begingroup$ Take $A$ to be the identity and $V$ the vector $[8, 8, 8, 8]^T$. Then $A^{-1}V = AV= V$. However, for $A$ eight times the identity and $V = [1, 1, 1, 1]^T$ you get a different $A^{-1}V$ (the one you computed). $\endgroup$ – Fabio Somenzi Jan 3 '18 at 4:32
  • $\begingroup$ This seems like an example of the XY problem. $\endgroup$ – user223391 Jan 3 '18 at 7:39
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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.

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  • $\begingroup$ It seems like V has the form $V=constant * [1,1,1,1]^T$. So, can I conclude that $A^{-1} V=c * [1,1,1,1]^T$? $\endgroup$ – Ying Jan 3 '18 at 5:59
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    $\begingroup$ No, unfortunately. I've just picked scaling matrices to illustrate how the eigenvector assumption doesn't guarantee you the solution. As another example, take $$A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}, V = \begin{bmatrix} 8 \\ 0 \\ 0 \\ 0 \end{bmatrix}.$$ $\endgroup$ – Theo Bendit Jan 3 '18 at 7:57
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    $\begingroup$ You're not directly assuming that $V$ is an eigenvector. You're assuming that $AV = [8, 8, 8, 8]^T$ and $V = [1, 1, 1, 1]^T$. Putting this together, we get, $AV = 8V$, which immediately implies $V$ is an eigenvector. $\endgroup$ – Theo Bendit Jan 4 '18 at 23:51
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    $\begingroup$ I don't mean it as a criticism of your solution: given the assumptions you've started with, your reasoning is perfectly valid. I was just pointing out that, underlying it all, what's allowing you to do this is the fact that your assumptions immediately imply that $V$ is an eigenvector of $A$ with a known eigenvalue. If you make other, seemingly analogous assumptions, e.g. $V = [1, 0, 0, 0]$, then you cannot proceed, as this means $V$ is not an eigenvector, and there will be many possibilities for $A^{-1}V$. $\endgroup$ – Theo Bendit Jan 4 '18 at 23:55
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    $\begingroup$ Thank you very much for your very detailed explanation! $\endgroup$ – Ying Jan 5 '18 at 17:06
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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.

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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 $V$.

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