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Consider a matrix $A= uv^T$ where $u=\begin{pmatrix}1 \\ 2 \end{pmatrix} , v = \begin{pmatrix}1 \\1 \end{pmatrix}$. Note that $v^T$ denotes the transpose of $v$. The largest eigenvalue of $A$ is ____
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$u = \begin{bmatrix} 1\\ 2 \end{bmatrix}$

$v^T= \begin{bmatrix} 1 & 1 \end{bmatrix}$

$A= uv^T$

$A=  \begin{bmatrix} 1\\ 2 \end{bmatrix}  \begin{bmatrix} 1 & 1 \end{bmatrix}$

$\quad =\begin{bmatrix} 1 &1 \\ 2 & 2 \end{bmatrix}$

$A - \lambda I = 0$

$\implies \begin{bmatrix} 1 - \lambda &1 \\ 2 & 2- \lambda \end{bmatrix} = 0$

$\implies  (1- \lambda) (2- \lambda) - 2 =0$

$\implies\lambda ^{2} - 3\lambda = 0$

$\implies \lambda = 0, 3$

So, maximum is $3$.
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8 votes
8 votes
after multiplying u and v Transpose
A=  1   1

      2    2

that means sum of eigen values will be 3 and their product will be 0
so the two eigen values will be 3 and 0

 

Answer will be 3
8 votes
8 votes

$Theorem$: If $\lambda$ is a non-zero eigen value of matrix $AB$, then same $\lambda$ is also an eigen value of matrix $BA$

$proof$:  

Let $\lambda \neq 0$ be a  eigen value of matrix $AB$

$\begin{align*} (AB)v &= \lambda v\\ B(AB)v&=B(\lambda v) \\ (BA)Bv &=\lambda (Bv) \end{align*}$

The point is that matrices $AB$ and $BA$ share non-zero eigen values


The dimension of $u$ and $v$ is $2$ x $1$.

The dimension of $A = uv^{T}$ is $2$ x $2$ and hence there will be two eigen values of $uv^{T}$ (repeating or non repeating)

The dimension of $v^{T}u$ is $1$ x $1$ 

$v^{T}u$ is dot product of $v$ and $u$ $ = 3$ and this implies eigen value of $v^{T}u$ is 3.

Since non-zero eigen values are shared by $uv^{T}$ and  $v^{T}u$

So eigen values of $uv^{T}$ are $3, 0$.

Largest of the eigen value is 3 which is nothing but dot product of $u$ and $v$

2 votes
2 votes
$u=\begin{bmatrix} 1\\ 2 \end{bmatrix} $  $v^{T}= \begin{bmatrix} 1 & 1 \end{bmatrix}$

$A= uv^{T}$

A= $\begin{bmatrix} 1 &1 \\ 2& 2 \end{bmatrix}$

Now you get equation

$(\lambda-1)(\lambda-2)- 2=0$

$\lambda ^{2}- 3\lambda=0$

$\lambda(\lambda -3)=0$

we get $\lambda =0,3$

answer will be $3$
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