For which $\mathbf X$ (matrix) does there exist a scalar $c$ such that $\mathbf{AX} = c\mathbf{X}$? $$ \mathbf{A} = \begin{bmatrix} 5 & 0 & 0\\ 1 & 5 & 0\\ 0 & 1 & 5 \end{bmatrix} $$ I have thought about the idea that $\mathbf{X}$ is $0$ so $c$ is $0$, but is it the only solution?

Thank you!


Write $X$ as a collection of columns, ie, $X = \begin{bmatrix} x_1 & \cdots & x_n \end{bmatrix}$. Then $AX = cX$ iff $Ax_i = c x_i$ for all $i$ iff $x_i \in \ker (A-cI)$ for all $i$.

In this particular case, since $5$ is the only eigenvalue, $\ker (A-cI) = \{0\}$ if $c\neq 5$ and $\ker (A-5I) = \mathbb{sp} \{e_3 \}$.

So all solutions are of the form $c \neq 5, X = 0$, or $c=5$ with $X= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \alpha_1 & \alpha_2 & \alpha_3 \end{bmatrix}$, for arbitrary constants $\alpha_i$.

Addendum: Here is an approach that doesn't use the eigenvalue/eigenvector language:

As above, write $X$ as a collection of columns. Then we still have $AX = cX$ iff $Ax_i = c x_i$ for all $i$, so we are interested in looking for solutions to the equation$Ax = cx$ or equivalently $(A-cI) x = 0$.

There are two cases to consider. If $c \neq 5$, the the matrix $A-cI$ is invertible, since it is lower triangular with non-zero diagonal elements. Consequently, if $(A-cI)x = 0$, then we must have $x=0$ (by multipliing both sides by the matrix $(A-cI)^{-1}$, or noting that the first row implies $x_1=0$, the second row then implies $x_2 = 0$ and finally the third row gives $x_3=0$).

Now consider the case $c = 5$, and consider $(A-5I)x = 0$. Notice that the second row implies $x_1=0$ and the third row implies $x_2=0$. Furthermore, $x_3$ can have any value whatsoever.

Combining the above, we see that $X$ must have the form $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \alpha_1 & \alpha_2 & \alpha_3 \end{bmatrix}$, for arbitrary constants $\alpha_i$.

  • $\begingroup$ i think i suppose to solve it in another way. i am doing this exercise from hoffman's book and i didn't learn about eigenvalues or something $\endgroup$ – TalBin Aug 28 '12 at 15:22
  • $\begingroup$ Let me add something to my answer. $\endgroup$ – copper.hat Aug 28 '12 at 15:24

No, $c = 5$ is the other solution, with $X$ given by $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ x_1 & x_2 & x_3 \end{pmatrix}$.

The trick is that the column vectors $v$ which make up $X$ satisfy $Av = cv$ and so are either eigenvectors or the zero vector.

The zero matrix $X = 0$ corresponds to any value of $c$ and not just $0$.

  • 1
    $\begingroup$ The columns of $X$ do not have to be identical... $\endgroup$ – copper.hat Aug 28 '12 at 14:45
  • $\begingroup$ @copper.hat Whoops! Thanks. $\endgroup$ – Cocopuffs Aug 28 '12 at 15:04

Note that your matrix is in Jordan's canonical form (normally the 1s are above the diagonal, but this form is also sometimes found in textbooks) and you can read out of the matrix that the only eigenvalue is 5, with geometric multiplicty 1.


Observe that the matrix has $5$ as the only eigenvalue, with algebraic multiplicity $3$. The corresponding eigenspace is

$$\langle x\rangle, \mbox{where } x = \begin{pmatrix}0\\0\\1\end{pmatrix}$$

so that $Ax=5x$. If you take the matrix $X=(x|x|x)$, you get


You can also take scalar multiples of $X$, but that's about it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.