GATE CSE 2018 | Question: 17

Question

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 ____

Comments

@shravan2001 first of all u can't use row/ col operations of Given MAtrix u've to Use it on $(A- \lambda I)$ on same  given matrix u'll get wrong answer .

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A  will be  - 

Here R2= 2R1
As rows are Linearly Dependent, Det(A)= 0 , thus,  1  eigen value is 0

Trace of Matrix = Sum of Eigen values

Therefore, 1+2 = 3= Sum of Eigen values,

As one of them is 0, other one has to be 3.
Hence, answer is 3.

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@shravan2001 

if a matrix is given u cannot change the values by row/ col operations and then find the eigen values .

those are not the eigen values of the original matrix.

Answer 1

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

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

$A= uv^T$

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

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

$A - \lambda I = 0$

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

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

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

$\implies \lambda = 0, 3$

So, maximum is $3$.

Answer 2

$Theorem$: 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

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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 $v$

Answer 3

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

Answer 4

$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 $3$

Comments

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

Answer 5

Q)

$$

\begin{aligned}

& A=u v^{\top} \text { (given) } \quad U=\left[\begin{array}{l}

1 \\

2

\end{array}\right] \quad V=\left[\begin{array}{l}

1 \\

1

\end{array}\right] \\

& A=\left[\begin{array}{l}

1 \\

2

\end{array}\right]_{2 \times 1}\left[\begin{array}{ll}

1 & 1

\end{array}\right]_{1 \times 2}=\left[\begin{array}{ll}

1 & 1 \\

2 & 2

\end{array}\right]_{2 \times 2}

\end{aligned}

$$

Now, $|A-\lambda I|=0$

$$

\begin{gathered}

\left|\begin{array}{cc}

1-\lambda & 1 \\

2 & 2-\lambda

\end{array}\right|=0 \\

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

\not2-\lambda-2 \lambda+\lambda^2-\not 2=0 \\

\lambda^2-3 \lambda=0 \\

\lambda=3 ; \lambda=0

\end{gathered}

$$

Langest eigevalue is $\lambda=3$

Answer 6

Here if know one note that : "If we have two vectors of same direction let's say A and B than AB and BA shares the non zero eigen values." (means the non-zero eigen values of AB and BA will always be same.)

So here if we use this property than -

A = uv^T , than the non-zero eigen values of uv^T and v^Tu will be same so instead of finding eigen values for uv^T, we can find for v^Tu and,

v^Tu = [1 1] [1       

                   2]                 

than    "v^Tu = [3] so "3" will be the only non-zero eigen value possible for both  and it is the maximum eigen value possible."