Relationship between a matrix with specific eigenvectors and eigenvalues

Question

If $A$ is a matrix for which its eigenvectors are only multiples of $\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$, then in general, can it be said that $A$ has repeated eigenvalues? Also, this matrix $A$ cannot be diagonalizable?

Answer 1

Since your vector is $3\times 1$, the matrix is $3\times 3$. For it to be diagonalizable, there have to be three linearly independent eigenvectors. Since every eigenvector is a multiple of the single vector, the matrix cannot have three linearly independent eigenvectors, so your matrix is definitely not diagonalizable.

In addition, to every distinct eigenvalue there is at least one eigenvector, and eigenvectors associated to distinct eigenvalues are linearly independent. So your matrix can only have a single eigenvalue.

If you are working over a non-algebraically closed field (like the reals), it could mean that there is only one eigenvalue with multiplicity $1$, but that the other factor of the characteristic polynomial is irreducible quadratic. Otherwise, or over the complex numbers, the eigenvalue is repeated $3$ times.

For an example of the latter, consider the matrix $$A=\left(\begin{array}{ccr} 1 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{array}\right).$$ Over the reals, the only eigenvalue is $1$ (the characteristic polynomial is $-(x-1)(x^2+1)$), which is not repeated. All eigenvectors are multiples of $\mathbf{e}_1$.

So $A$ is definitely not diagonalizable, but it could have a single eigenvalue of algebraic multiplicity $1$. Otherwise, it has a single eigenvalue of algebraic multiplicity $3$.

Answer 2

If you want a better understanding about eigenvectors, or start from eigenvectors and eigenvalues and then reason about the possible transformations they could belong to, eigencircles are a good tool. They were introduced/defined in 2006 by Englefield and Farr. Too bad eigencircles only work for 2x2 matrices.

Set $A=\left[\begin{matrix}a&b\\c&d\\\end{matrix}\right]$, $A$ is the matrix of the linear transformation $\mathfrak{t}$

The reasoning about eigencircle starts from the observation that if you apply a linear transformation to 'a' vector $\vec(x)$, the 'displacement' from $\vec(x)$ to $\mathfrak{t}(\vec{x})$ can always be described by a rotation and a change of length of the original vector.
The eigencircle of a transformation $\mathfrak{t}$ or a matrix $A$ is the collection of all possible rotations and scalings 'produced' by the linear transformation.

The eigencircle can be described both in a polar and Cartesian form, that is why I give the different formulations with ${EC(\mathfrak{t})}_{cart}$ and ${EC(\mathfrak{t})}_{polar}$.

The eigencircle ${EC(\mathfrak{t})}_{cart}$ or ${EC(\mathfrak{t})}_{polar}$ of a linear transformation $\mathfrak{t}$ is a circle defined by all possible tuples $(s,\theta)$
where
$\theta=\angle\left(\vec{x,}\mathfrak{t}\left(\vec{x}\right)\right)$ is the angle between an original vector $\vec{x}$ and its image $\mathfrak{t}\left(\vec{x}\right)$ and
$s=\frac{\|\mathfrak{t}(\vec{x})\|}{\|\vec{x}\|}$ is the scaling $s$ from the original length $\|\vec{x}\|$ to the final length $\|\mathfrak{t}(\vec{x})\|$.

${EC(\mathfrak{t})}_{polar}=\left\{\left(\ s_{\vec{x}},\theta_{\vec{x}}\right)| \exists\vec{x}=\left[\begin{matrix}x\\y\\\end{matrix}\right]\ and\ \mathfrak{t}\left(\vec{x}\right)=\left[\begin{matrix}s_{\vec{x}}&0\\0&s_{\vec{x}}\\\end{matrix}\right]\left[\begin{matrix}\cos{\theta_{\vec{x}}}&-\sin{\theta_{\vec{x}}}\\+\sin{\theta_{\vec{x}}}&\cos{\theta_{\vec{x}}}\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=A\left[\begin{matrix}x\\y\\\end{matrix}\right]\right\}$

${EC(\mathfrak{t})}_{polar}=\left\{\left(s_{\vec{x}},\theta_{\vec{x}}\right) |\ \exists \vec{x}\ such\ that\ s_{\vec{x}}=\frac{\|\mathfrak{t}(\vec{x})\|}{\|\vec{x}\|} \ and\ \theta_{\vec{x}}=\angle(\vec{x},\mathfrak{t}(\vec{x}))\right\}$

${EC(\mathfrak{t})}_{cart}=\left\{\left(\lambda,\mu\right)|\exists\vec{x}=\left[\begin{matrix}x\\y\\\end{matrix}\right]and\ \mathfrak{t}\left(\vec{x}\right)=\left[\begin{matrix}\lambda &-\mu\\+\mu&\lambda\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}s_{\vec{x}}&0\\0&s_{\vec{x}}\\\end{matrix}\right]\left[\begin{matrix}\cos{\theta_{\vec{x}}}&-\sin{\theta_{\vec{x}}}\\+\sin{\theta_{\vec{x}}}&\cos{\theta_{\vec{x}}}\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]\right\}$

Normally you start from a known transformation and look for its properties.
You know $\mathfrak{t}$ and visualize its properties.
But you can reason backwards too.
The eigenvectors and eigenvalues are determined by the points $(\lambda_{A1},0)$ and $(\lambda_{A2},0)$ and $G(a,c)$.
So the set of all matrices with the same eigenvalues is the set of all circles containing the points $(\lambda_{A1},0)$ and $(\lambda_{A2},0)$.

For completeness I add the definitions you need to construct an eigencircle: $$f=\frac{\left(a+d\right)}{2}$$ $$g=\frac{\left(c-b\right)}{2}=-\frac{\left(b-c\right)}{2}$$ I explicitly state $-(b-c)$ since 'my' formulas are different from the ones in the articles of Englefield and Farr. $$r^2=f^2+g^2$$ $$\det{\left(A\right)}=r^2-\rho^2$$ $$\rho^2=\left(\frac{a-d}{2}\right)^2+\left(\frac{b+c}{2}\right)^2$$ $$\left(\lambda-f\right)^2+\left(\mu-g\right)^2-\rho^2=0 \ \ (1)$$ (1) is the equation of the eigencircle of the matrix $A$

If you need more detail, just ask it in a comment and I will add some more detail.
You can also visit www.heavisidesdinner.com.
It shows the complete derivation and some geogebra examples.

There is a Geogebra example you can play with, at:
Public Geogebra example showing eigencircles of $A$, $A^T$, $A^{-1}$

$A$ ">