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

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.

• Just take the result in the linked question and add $\lambda$ to all eigenvalues (the eigenvectors are the same). – Hans Lundmark Jan 4 '18 at 20:26
• $\mathbf v^T\mathbf v=\|\mathbf v\|^2$ – amd Jan 5 '18 at 1:10
• Hint: for eigenvalue problems, addition of $\lambda\mathbf I$ just shifts the eigenvalues by$~\lambda$ (otherwise same eigenvectors), so you can first consider the problem without that term. Doing so, this question asks about eigenvectors of a rank$~1$ matrix, which is easy. – Marc van Leeuwen Jan 6 '18 at 10:19

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.

Hint: Consider a vector orthogonal to $v$.

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 $\lambda$.

• Why the first part of LHS $\vec{v}(\vec{v}^T\vec{w}_i) + \lambda \vec{w}_i = \lambda \vec{w}_i$ is gone? – Ying Jan 4 '18 at 19:01
• @Ying Since each $\mathbb{w}_i$ is orthogonal to $\mathbb{v}$ their dot product ($\mathbb{v}^T\mathbb{w}_i$) is zero. – Ryan T Johnson Jan 4 '18 at 19:08
• Thanks. Do we know what's form look like for $\mathbf{w}_i$? – Ying Jan 4 '18 at 19:15
• @Ying There are infinite choices for the $\mathbf{w}_i$, however, their span is equal to $E_{\lambda} = \{ \left <x_1,...,x_n\right> \; | \; x_1v_1 + x_2v_2 + ... x_n v_n = 0 \}$ where $\left<v_1,...,v_n\right> = \mathbf{v}$. – Ryan T Johnson Jan 4 '18 at 20:06