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?

 A: 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. 
A: 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. 
A: My answer is for a part; let the order of matrix be n then we know the very common relation n-r=1 it implies r=n-1 hence rank is less than n hence not invertible 
