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

Eigen values x=0,3 but maximum  value is 3

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largest eigen value is 3.

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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 1

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 2

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."