Linear Algebra

We know, cholesky decomposition,
A=LL^{T}
Where, L= lower tringular matrix
\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{cc}L_{11} & 0 \\ L_{21} & L_{22}\end{array}\right]\left[\begin{array}{cc}\mathrm{L}_{11} & L_{21} \\ 0 & L_{22}\end{array}\right]

\left[\begin{array}{cc}8 & -5 \\ -5 & \mathrm{a}_{22}\end{array}\right]=\left[\begin{array}{cc}\mathrm{L}_{11}^{2} & \mathrm{~L}_{11} \mathrm{~L}_{21} \\ \mathrm{~L}_{11} \mathrm{~L}_{21} & \mathrm{~L}_{21}^{2}+\mathrm{L}_{22}^{2}\end{array}\right]
On comparison on both sides,
\mathrm{L}_{11}=\sqrt{8}=2 \sqrt{2}
and,\;\; L_{11} L_{21}=-5
\mathrm{L}_{21}=\frac{-5}{2 \sqrt{2}}
and,\;\; a_{22}=L_{21}^{2}+L_{22}^{2}
=\left(\frac{-5}{2 \sqrt{2}}\right)^{2}+1.968^{2}
=6.998 \approx 7

|A|=\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|
=1(2-1)+1(-1)+0
=1-1=0

So A X=B does not have unique solution because \rho(A) \lt 3
|A-\lambda|=0 \Rightarrow\left|\begin{array}{ccc}1-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 1-\lambda\end{array}\right|=0
(1-\lambda)((2-\lambda)(1-\lambda)-1)+1(-1+\lambda)+0=0
(1-\lambda)\left(\lambda^{2}-3 \lambda+1\right)-1+\lambda=0
(1-\lambda)\left(\lambda^{2}-3 \lambda+1-1\right)=0
(\lambda-1)\left(\lambda^{2}-3 \lambda\right)=0
\lambda=0,1,3

Matrix A has three distinct Eigen values so have three linearly independent eigen vectors. so option (C) is correct.
Given matrix is symmetric matrix with real value entries. Hence A is not a positive definite matrix. because
|1|=1 \gt 0
\left|\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right|=2-1=1>0
\left|\begin{array}{ccc}1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1\end{array}\right|=0 (which is not positive)
Hence option (D) is incorrect.

A x=\lambda x \ldots (i)
A^{T} x=\lambda x \ldots (ii)
A and A^T both have same eigen values and eigen vectors.
A x=\lambda x \ldots(i)
\Rightarrow \quad A^{-1} A x=A^{-1}(\lambda x)=\lambda A^{-1} x
\Rightarrow \quad x=\lambda A^{-1} x
A^{-1} x=\frac{1}{\lambda} x
So, eigen value and eigen vector of A^{-1} is \frac{1}{\lambda} x and \mathrm{x}.

M X=\lambda X
[\lambda is eigen value of (M)]
\begin{aligned} M M X & =\lambda M X \\ M^{2} X & =\lambda(\lambda X) \\ M^{2} X & =\lambda^{2} X \end{aligned}

\left[\lambda^{2}\right. (non negative) is eigen value of \left.\mathrm{M}^{2}\right]
\begin{aligned} M X & =\lambda X \\ M^{\top} X & =\lambda X \end{aligned}

\left[M, M^{\top}\right. have same eigen values]
\begin{aligned} & M M^{\top} X=\lambda M X \\ & M M^{\top} X=\lambda(\lambda X) \\ & M M^{\top} X=\lambda^{2} X \end{aligned}

\left[\lambda^{2}\right. (none negative) is eigen value of \left.M M^{\top}\right]
\begin{aligned} M X & =\lambda X \\ M^{\top} X & =\lambda X \\ M^{\top} M X & =\lambda M X \\ M^{\top} M X & =\lambda(\lambda X) \\ M^{\top} M X & =\lambda^{2} X \end{aligned}

\left[\lambda^{2}\right. (non negative) is eigen value of \left.M^{\top} M\right]
\begin{aligned} M X & =\lambda X \\ M M X & =\lambda M X \\ M^{2} X & =\lambda(\lambda X) (M^T)^2 X&= \lambda ^2 X \end{aligned}
\left[\because M, M^{\top}\right. have same eigen value]
\left(M^{\top}\right)^{2} X=\lambda^{2} X
[ \lambda^{2} is eigen value of \left(M^{\top}\right)^{2} which non negative] Hence, option A, B, C, D are correct.

Let vector
\begin{aligned} y&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}_{3 \times 1}\\ y^T&=\begin{bmatrix} 4& 4& 4 \end{bmatrix}_{1 \times 3}\\ yy^T&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}\begin{bmatrix} 4& 4& 4 \end{bmatrix}\\ &=\begin{bmatrix} 16 & 16 & 16\\ 16 & 16 & 16\\ 16 & 16 & 16 \end{bmatrix}\\ y^Ty&=[4^2+4^2+4^2]_{1 \times 1}\\ &=[48]_{1 \times 1}\\ \rho (y)&=\rho (y^T)=\rho (yy^T)=\rho (y^Ty)=1 \end{aligned}
From above information
yy^T is asymmetric.
y^Ty is an eigen value of yy^T .
yy^T has rank 1
det(yy^T) =0 so, yy^T is not invertible.