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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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Eigen values x=0,3 but maximum  value is 3
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largest eigen value is 3.

u = $\begin{bmatrix} 1\\ 2 \end{bmatrix}$

vT= $\begin{bmatrix} 1 & 1 \end{bmatrix}$

A= uvT

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

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

A - λ I = 0

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

=>  (1- λ) (2- λ) - 2 =0

=> $\lambda ^{2} - 3\lambda = 0$

=> λ = 0, $3$

So maximum is $3$.

edited by
+1 vote
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

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

$A= uv^{T}$

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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sir, last entry of A, should be 2 but not 4

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