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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 ____
asked in Linear Algebra by Boss (16k points)
edited by | 2.5k views
+1
Answer is 3
0
Eigen values x=0,3 but maximum  value is 3
+3
largest eigen value is 3.

4 Answers

+10 votes
Best answer
$u = \begin{bmatrix} 1\\ 2 \end{bmatrix}$

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

$A= uv^T$

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

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

$A - \lambda I = 0$

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

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

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

$\implies \lambda = 0, 3$

So, maximum is $3$.
answered by Loyal (9.1k points)
edited by
+2 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
answered by (283 points)
+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 $3$
answered by Veteran (61.4k points)
edited by
0
sir, last entry of A, should be 2 but not 4
0 votes
answer is 3
answered by (71 points)
Answer:

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