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Find the eigenvalues and eigenvectors of the matrix $A = uu^t$, where $u\in\mathbb{R}^n$

The multiplication will give me a $n \times n$ matrix like this:

$$\begin{bmatrix} u_1^2 & u_1 u_2 & \dots & u_1u_n \\ u_2 u_1 & u_2^2 & \dots & u_2u_n \\ \vdots & \vdots & \ddots & \vdots \\ u_n u_1 & \dots & \dots& u_n^2 \end{bmatrix}$$

I suppose there is some trick using the fact that this matrix is symmetric and square. This should help taking the determinant

$$\det \begin{bmatrix} u_1^2 - \lambda & u_1 u_2 & \dots & u_1u_n \\ u_2 u_1 & u_2^2 - \lambda & \dots & u_2u_n \\ \vdots & \vdots & \ddots & \vdots \\ u_n u_1 & \dots & \dots& u_n^2 - \lambda \end{bmatrix}$$

Rodrigo de Azevedo
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Poperton
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  • See this answer. The only difference is that $|u|$ might not be equal to $1$, so the lone nonzero eigenvalue might not be $1$, either. – amd Aug 09 '18 at 04:16

4 Answers4

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$$A = uu^T$$ Notice that $$Au = uu^Tu = u(u^Tu) = (u^Tu) u = \lambda u $$ where $$\lambda = u^Tu$$ is an eigenvalue corresponding to eigenvector $u$. All other eigenvalues are zero because $A$ is an outer-product of $u$ on itself.

Why zero-eigenvalues? Consider vectors of the form $$v = (I - \frac{1}{u^Tu}uu^T)\alpha$$ then $$Av =uu^Tv = uu^T\alpha - \frac{1}{u^Tu}uu^Tuu^T\alpha =uu^T\alpha - \frac{1}{u^Tu}(u^Tu)uu^T\alpha = uu^T\alpha-uu^T\alpha = 0$$ Turns out there are $n-1$ linearly and independent vectors $v_1 \ldots v_{n-1}$ that satisfy the above equation. So $0$ has $n-1$ multiplicity.

Ahmad Bazzi
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To elaborate on what Ahmad didn't mention is that if you take the outer product of a vector with itself you'll end up with a rank one matrix because all the columns are linearly dependent on each other. If the rank of the matrix is only 1 you see the rest of the eigenvalues are zero.

This is detailed here.

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Added: Observe that $x^tAx=x^t(uu^t)x=(u^tx)^t(u^tx)=\vert\vert u^tx\vert\vert^2\ge 0$ for any $x\in\mathbb R^n$ .

<p>Thus matrix $A$ is semi-positive definite. Also $Rank(A)=1$ and $A$ is symmetric then $1=p+n$ (where $p$ and $n$ respectively denote the number of positive and negative eigenvalues) gives $p=1$.</p>

<p>Using $trace(A)=$sum of eigenvalues you get $\lambda=u_1^2+u_2^2+....+u_n^2$ as the only non-zero eigenvalue and rest $(n-1)$ eigenvalues are zero.</p>

Another way is to consider $u$ as a $1×n$ row vector which gives you $A=[u_1^2+u_2^2+....+u_n^2]$. Then $Ax=\lambda x$ is trivially satisfied for all $x\in \mathbb R$ thereby giving $\lambda=u_1^2+u_2^2+....+u_n^2$ as eigenvalue and $x$ as eigenvector.

Nitin Uniyal
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$uu^T$ is the matrix of the endomorphism $x\mapsto u\langle u,x\rangle$ expressed in the canonical basis of $R^n.$ Assuming $u\neq 0$ this endomorphism expressed in the orthonormal basis $(\frac{u}{\|u\|}, f_2,\ldots,f_n)$ has matrix $$\mathrm{diag}(\|u\|^2,0,0,\ldots,0)$$ and everything becomes trivial.

Gérard Letac
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