# What if the only eigenvectors of $A$ are multiples of $x=(1,0,0)^T$?

The question is from an exercise in Gilbert Strang's Linear Algebra and its Applications:

Suppose the only eigenvectors of $A$ are multiples of $x=(1,0,0)$. True or false:

(a) $A$ is not invertible.

(b) $A$ has a repeated eigenvalue.

(c) $A$ is not diagonalizable.

(b) has to be true. Since if $A$ has different eigenvalues, then there will be linear independent eigenvectors.

One example that I can come up with for which (a) and (c) are true is $$A=\left( \begin{array}{ccc} 0& 1& 0 \\ 0& 0 & 1 \\ 0& 0 & 0 \\ \end{array} \right)$$

Here are my questions:

• Are there any counterexamples for (a) and (c)?
• What's the underlying picture of this problem?
-
The underlying picture is the theory of Jordan normal form. Are you familiar with it? –  Qiaochu Yuan Aug 2 '11 at 20:21
If your field is not algebraically closed, it may well be the case that the only eigenvectors are the non-zero multiples of $(1,0,0)$ and there is no repeated eigenvalue: for example, over $\mathbb R$ the characteristic polynomial could be $x(x^2+1)$. –  Mariano Suárez-Alvarez Aug 2 '11 at 20:22
I don't understand... you have correctly pointed out (b) must be true, so how could you have a counterexample to (b), let alone to (a) and (b)? –  Arturo Magidin Aug 2 '11 at 20:22
@Arturo: Ah, that's my silly typo. Corrected. Thanks. –  Jack Aug 2 '11 at 20:24
@Jack: There can be no counterexample to (c), because if the matrix is diagonalizable then it necessarily must have a basis of eigenvectors, hence at least three linearly independent eigenvectors, hence not every eigenvector is a scalar multiple of $(1,0,0)$. –  Arturo Magidin Aug 2 '11 at 20:29

Over the complex numbers, you must have a matrix that is similar to $$\left(\begin{array}{ccc} \lambda & 1 & 0\\ 0 & \lambda & 1\\ 0 & 0 & \lambda \end{array}\right),$$ where $\lambda$ is the unique eigenvalue; this follows form the theory of Jordan Canonical forms.

Over a non-algebraically closed field, you have two possibilities: either a matrix similar to one as above (if the characteristic polynomial splits as a perfect cube), or else a matrix similar to one of the form $$\left(\begin{array}{ccr} \lambda & 0 & 0\\ 0 & 0 & -a\\ 0 & 1 & -b \end{array}\right),$$ where $\lambda$ is the eigenvalue, and $x^2 + bx + a$ is an irreducible quadratic polynomial over your field; this follows from the theory of Rational Canonical Forms.

In either case, the matrix is invertible if and only if $\lambda\neq 0$, and the matrix is never diagonalizable.

-
+1. Thanks for you answer. I haven't learned the Rational Canonical Forms before. I guess it can be found in some advanced linear algebra book? –  Jack Aug 2 '11 at 20:35
@Jack: Well, it's in Friedberg, Insel, and Spence, at least. –  Arturo Magidin Aug 2 '11 at 20:36
Also in Artin! (Ch. 12) –  Ben Blum-Smith Aug 5 '11 at 11:28

In my elementary linear algebra class we did many problems like this without the Jordan Normal Form, etc, which I picked up later. So I will try to give a simpler explanation.

a) Recall that a matrix is non-invertible if it has nonzero vectors in its null space. This also means that those vectors are eigenvectors corresponding to the eigenvalue $0$. So what if $\text{null } A =\text{span }{(1,0,0)}?$ This is the case where an eigenvalue of $A$ is 0. However, you CANNOT conclude this from your givens, because you have no hypothesis on what the eigenvalue of $\text{span }{(1,0,0)}$ is.

b) Now, $A$ is a linear operator on $\mathbb R^3.$ This means that $A$ has a real eigenvalue, because the characteristic equation of $A$ is a cubic, so it has a real root. What if two of those roots were imaginary? Then $A$ would have its one-dimensional eigenspace, as given, but would not have a repeated eigenvalue.

c) This one is true. A matrix is diagonalizable if and only if has $n$ linearly independent eigenvectors, where $n$ is the dimension of the vector space it operates on. This means you need $3$ linearly independent eigenvectors where you only have one! An easy to remember this is that, to diagonalize a matrix $A$, you need to be able to write it as $A = PDP^{-1},$ where $P$ is an $n$ x $n$ matrix whose columns consist of eigenvectors of $A.$ Remember that $P$ is invertible if and only if it has linearly independent columns, and you are done.

-