# Find $X$ such that there exists a scalar $c$.

Let $A = \begin{bmatrix}5 & 0 & 0 \\1 & 5 & 0 \\ 0 & 1 & 5\end{bmatrix}$. For which $X$ does there exist a scalar $c$ such that $AX = cX$? Do they mean $X$ is the null matrix?

• They probably mean that $x$ is a vector and not the zero vector. This is called an eigenvector, and the scalar $c$ is called an eigenvalue. – TravisJ Mar 23 '15 at 1:58
• Or if $X$ is allowed to be a larger matrix, then every column of $X$ must be an eigenvector of $A$ (or the zero vector). Moreover, all the columns must correspond to the same eigenvalue, though in this example that doesn't matter, as there is only one eigenvalue. – David Mar 23 '15 at 2:05

The question is asking about a very important concept in linear algebra called eigenvalues (and eigenvectors). Eigen is the German word for 'proper', and by definition, an eigenpair is a non-zero vector $X$ and a scalar $c$ (which can be zero) that solves the equation $AX = cX$. Finding eigenpairs is a huge part of linear algebra, and if you haven't learned about it yet, you probably will soon.
In general for a 3x3 matrix the method is pretty simple. You rewrite the equation as $AX - cX = 0$, and then factor out $X$ to get $(A - cI)X = 0$ (where $I$ is the 3x3 identity, so $cI$ is just a matrix with all zeros except $c$'s on the diagonal). Since $X$ is nonzero, you now need to find the values of $c$ for which $A-cI$ has a nontrivial null space, which is often done by finding $c$ that make $\det (A-cI) = 0$. This boils down to solving a cubic polynomial equation.
In your case this is quite easy: $\det(A-cI) = (5-c)^3$, so the only eigenvalue is $c = 5$. If you want to check that this works, try to find the null space of $A-cI$. It should be three dimensional.