GO Classes 2024 | Weekly Quiz 7 | Linear Algebra | Question: 9

Answer 1

Statement 1:

we know $Ax= \lambda x$ , where $x $ is the eigenvector corresponding to eigenvalue $\lambda $

As $A$ is invertible then $A^{-1}x = \lambda^{-1}x$ so stmt 1 is $True$

Statement 2:

we know that eigenvalues of skew-symmetric matrix(i.e. $A^{T} = – A$) are pure imaginary

$\lambda = \pm ib$ ,where b is some  constant and they are always in pair

so if $\lambda  = ib$ is an eigen value then  $\bar{\lambda} = – ib = -\lambda$ is also be the eigenvalue

Hence stmt 2 is also $True$

Answer 2

Statement-1

we are given $Ax= \lambda x$ and we have to prove $A^{-1}x= \frac{1}{\lambda}x$

Multiply by $A^{-1}$ on both sides.

$A^{-1}Ax= \lambda A^{-1}x \Rightarrow$ $Ix= \lambda A^{-1}x$

$\mathbf{ \frac{1}{\lambda}x=A^{-1}x}$, which tells us that $\frac{1}{\lambda}$ is eigen value of $A^{-1}$

Therefore Statement-1 is True.

Statement-2

we are given that $A^T=-A$ and $Ax= \lambda x$

Multiply by $x^T$ on both sides

$x^{T}Ax=\lambda x^{T}x \Rightarrow x^{T}Ax=\lambda |x|^{2}$

We know that $a^Tb=b^Ta$

$x^{T}(Ax) \Rightarrow (Ax)^Tx \Rightarrow x^TA^Tx \Rightarrow -x^TAx$ chip in the value of $Ax$ form first line.

we get $x^{T}(Ax) = -x^T\lambda x $

$x^{T}(Ax) = –\lambda x^T x$ 

$x^{T}(Ax) = –\lambda |x|^2$

But $x^{T}(Ax)=\lambda |x|^2$

which means $\lambda |x|^2=-\lambda |x|^2 \Rightarrow \lambda =-\lambda$

source

Therefore Statement-2 is True. 

Comments

@rhl what is a and b here?

---

a and b are column vectors

---

ok, my doubt is in 4^th line of explanation of statement 2 did you take transpose of $a^Tb$?

---

No, this is a property i am using. $a^Tb=b^Ta$ (dot product of 2 vectors.).

Check the next statement in $x^TAx$ , $x^T$ is $a^T$ and $Ax$ is $b$ and then i am applying the above property.

Answer 3

The complete proof of statement 2

A is a skew symmetric matrix

$A' = -A$

Let $x$ be an eigen vector of $A$ with corresponding eigen value $\lambda$

$Ax=\lambda x$

Apply conjugate on both sides

$\begin{align*} \overline{Ax} &=\overline{\lambda x} \\ \bar{A}\bar{x} &=\bar{\lambda}\bar{x} \\ A\bar{x} &= \bar{\lambda}\bar{x} --- eq(1) \end{align*}$

$Ax = λx$

Left multiply by the transpose of $\overline{x}$ on both sides

$\begin{align*} (\bar{x})'(Ax) &=(\bar{x})'\lambda x \\ (A'\bar{x})'x &=\lambda (\bar{x})'x \\ (-A\bar{x})'x &=\lambda (\bar{x})'x \end{align*}$

Using $eq1$

$\begin{align*} {(-A\overline{x})}'x &= \lambda {(\overline{x})}'x\\ {(-\overline{\lambda}\overline{x})}'x &= \lambda {(\overline{x})}'x \\ -\overline{\lambda}{(\overline{x})}'x &= \lambda {(\overline{x})}'x \\ -\overline{\lambda}{(\overline{x})}'x - \lambda {(\overline{x})}'x &= 0\\ (-\overline{\lambda}-\lambda) {(\overline{x})}'x &= 0 \end{align*}$

Since by definition Eigen vectors are not zero, 

$\begin{align*} (-\overline{\lambda}-\lambda) &= 0 \end{align*}$

$\begin{align*} \lambda &= -\overline{\lambda} \end{align*}$

Let $λ$ be $a+ib$

then – $\overline{λ} = -a+ib$

$a+ib = -a+ib$

So $a = 0$

So  $λ$ is purely imaginary or $0$

We have proved an important result here that the eigen values of skew-symmetric matrices is 0 or purely imaginary

The only skew-symmetric matrix that satisfies λ = 0 is the zero matrix

For non-zero skew-symmetric matrices, eigen values are purely imaginary

For any matrix A if $\lambda$ is eigen value corresponding to vector $v$, then $\overline{\lambda}$ is eigen value corresponding to vector $\overline{v}$

$\begin{align*} \lambda &= ib\\ \overline{\lambda} &= -ib = -\lambda \end{align*}$

Hence the proof.

Some corollary/results from this proof

1. The Eigen value of skew symmetric matrices is either $0$ or pure imaginary
    
2. The eigenvalues of non zero skew-symmetric matrices come in pair of
     $\lambda$ and $-\lambda$ for eigen vectors  $v$ and $\overline{v}$ respectively
    
3. The eigenvalues of skew-symmetric matrices λ and -λ have the same Algebraic multiplicities – this is extra info

Comments

how come  −$\lambda$ conjugate = -ib  if $\lambda$=ib
 

as if $\lambda$=ib  then -$\lambda$ = -ib  and conjugateOf(-$\lambda$) = ib

---

It is $\lambda = -(\overline{\lambda})$ as eigen values are purely imaginary in case if non-zero skew-symmetric matrices.
Sorry for the incomplete proof