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Given a nonzero column vector $A$=$[a_1 a_2.......a_n]^T$. Find the non zero eigen values and eigen vectors for $A$$A^T$.

I have no idea.what theorem should I apply or what I have to do to solve this. I know that fact $A$$A^T$ and $A^T$$A$ have the same eigen values and if $A$ and $B$ be square matrix of same order the $AB$ and $BA$ have the same eigen values. Is it related to this statements?

Sunit das
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    I suggest you take an example. Take $A = [1,2]^T$. What is $AA^T$? What are its non-zero eigenvalues and eigenvectors? Now change $A$ to something else and check : find a pattern, and see if you can generalize. – Sarvesh Ravichandran Iyer Oct 12 '20 at 05:38

2 Answers2

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$\newcommand{\rank}{\mathrm{rank}}$

This is a well known exercise in eigenvalue-eigenvector, for which you actually do not need to start with solving the equation $\det(\lambda I - B) = 0$ to determine eigenvalues (as you usually do). The special structure of $AA^T$ as a product of two rank one vectors allows us to have the following shortcut.

Since $\rank(A) = 1$, $\rank(AA^T) = \rank(A) = 1$, hence $AA^T$ has one and only one non-zero eigenvalue, and all the remaining $n - 1$ eigenvalues are $0$.

Now note $A^TA = a_1^2 + a_2^2 + \cdots + a_n^2 \neq 0$ is a scalar, denoted it by $\lambda$, hence \begin{align*} (AA^T)A = AA^TA = (A^TA)A = \lambda A. \end{align*} This shows $\lambda$ is an eigenvalue of $AA^T$, and $A$ is the eigenvector associated with $\lambda$.

Zhanxiong
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  • Thank you for this proof its like perfect I could have never came up with that. But as Teresa Lisbon suggested the pattern was clear to me. – Sunit das Oct 12 '20 at 09:26
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Suppose $\boldsymbol{v},\boldsymbol{w}\in\Bbb R^n$ are represented as column vectors. Recall that the dot product of $\boldsymbol{v}$ and $\boldsymbol{w}$ may be written as $\boldsymbol{v}\cdot\boldsymbol{w}=\boldsymbol{v}^\intercal\boldsymbol{w}$.

Your matrix is of the form $M=\boldsymbol{v}\boldsymbol{v}^\intercal$ where $\boldsymbol{v}\in\Bbb R^n$.

First, note that $$ M\boldsymbol{v}=\boldsymbol{v}\boldsymbol{v}^\intercal\boldsymbol{v}=\lVert\boldsymbol{v}\rVert^2\cdot\boldsymbol{v}\tag{1} $$ Here, we have used the identity $\boldsymbol{v}^\intercal\boldsymbol{v}=\boldsymbol{v}\cdot\boldsymbol{v}=\lVert\boldsymbol{v}\rVert^2$.

Next, suppose $\boldsymbol{w}$ is any vector orthogonal to $\boldsymbol{v}$, so that $\boldsymbol{v}\cdot\boldsymbol{w}=0$. Then $$ M\boldsymbol{w}=\boldsymbol{v}\boldsymbol{v}^\intercal\boldsymbol{w}=0\cdot\boldsymbol{v}=\boldsymbol{O}\tag{2} $$ Do you see how the equations (1) and (2) relate to the eigenvalue/eigenvector problem?

Brian Fitzpatrick
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