Eigenvector and eigenvalue of $\mathbf A+\lambda\mathbf I$ where $\mathbf A=\mathbf{vv}^\top$

Question

Given $\mathbf v=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}$ and $\mathbf A=\mathbf{vv}^\top$, find the eigenvectors and eigenvalues of $\mathbf A+\lambda\mathbf I$.

My current work progress is:

Since $(\mathbf A+\lambda\mathbf I)\mathbf v =\mathbf{Av}+\lambda\mathbf v=\mathbf v(\mathbf v^\top\mathbf v) + \lambda\mathbf v=\mathbf v \|\mathbf v\|+ \lambda\mathbf v=(\|\mathbf v\|+\lambda)\mathbf v$ ,

so one of the eigenvalue is $\lambda_1=(\|\mathbf v\|+\lambda)$ with corresponding eigenvector $\mathbf v_1=\mathbf v$.

I'd like to find out other pairs of eigenvalues and eigenvectors. I first calculate the trace.

$$\mathrm{Tr}\,(\mathbf A)=n\lambda+\|\mathbf v\|=\lambda_1+(n-1)\lambda$$

Hence, the rest of $n-1$ eigenvalues has sum $(n-1)\lambda$.

From here, I do not know how to proceed.

• Can I assume the rest of eigenvalues are $\lambda$ with multiplicities $n-1$? Why?
• How to find corresponding eigenvectors?

[Note1] This is like a follow up question regarding this one.

[Note2] Found later on there is a question but people focus on linking it to Note1.

Answer 1

For any matrix $A$, if $\mu$ is an eigenvalue of $A$, then $\mu+\lambda$ is an eigenvalue of $A+\lambda I$, with the same eigenvectors: $(A+\lambda I)\mathbf v = A\mathbf v+\lambda\mathbf v = \mu\mathbf v+\lambda\mathbf v=(\mu+\lambda)\mathbf v$. So, as Hans Ludmark commented, you just need to take the eigenvalues of $vv^T$ from the results in the linked questions and add $\lambda$ to them.

To recap, every column of $vv^T$ is a multiple of $v$, so its image is spanned by $v$. Thus, $v$ is an eigenvector, with eigenvalue $v^Tv$. The kernel of $vv^T$ is therefore $(n-1)$-dimensional, so the other eigenvalue is $0$, with multiplicity $n-1$. The kernel of $vv^T$ consists of vectors that satisfy $vv^Tw=(v^Tw)v=0$, therefore it’s the orthogonal complement of the span of $v$.

This means that the eigenvalues of $vv^T-\lambda I$ are $v^Tv+\lambda$, with multiplicity one, and $\lambda$, with multiplicity $n-1$. Any non-zero scalar multiple of $v$ is an eigenvector of the first eigenvalue, while any non-zero vector orthogonal to $v$ is an eigenvector of the second.

Answer 2

Hint: Consider a vector orthogonal to $v$.

Answer 3

Let $\mathbf{v}^{\perp}$ be the subspace of $\mathbb{R}^n$ of all vectors orthogonal to $\mathbf{v}$. (The proof that $\mathbf{v}^{\perp}$ is indeed a subspace is often proven or given as homework in most undergraduate linear algebra textbooks).

Let $\mathbf{w}_1, ..., \mathbf{w}_{n-1}$ be a basis of $\mathbf{v}^{\perp}$. Then by definition of orthogonal, $\mathbf{v}^T\mathbb{w_i} = 0$ for all $i$, so $$ (A + \lambda I)\mathbf{w}_i = \mathbf{v}(\mathbf{v}^T\mathbf{w}_i) + \lambda \mathbf{w}_i = \lambda \mathbf{w}_i $$

Thus $\{\mathbf{w}_1,...,\mathbf{w}_{n-1}\}$ are your other eigenvectors with eigenvalue $\lambda$.