How to calculate the rank of matrix with vertical and horizontal line

Question

I am sorry to post here, but I do not know this kind of matrix. Can someone tell what does the vertical and horizontal line mean and how can I calculate the rank of this matrix?:

$$A=\left[\begin{array}{ccc|ccc}1&1&-1&0&0&0\\2&-3&0&0&0&0\\1&0&1&0&0&0\\\hline 0&0&0&2&1&2\\0&0&0&-2&0&1\\0&0&0&3&2&1\end{array}\right]$$

Thank you very much!

Answer 1

It is a block-diagonal matrix, hence its determinant is the product of the determinants of the diagonal blocks.

Answer 2

Your matrix is a block matrix. The vertical and horizontal lines mean blocks. You can separate the matrix like this

$$ A= \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \tag{1}$$

following from another post

$$M=\underbrace{\begin{bmatrix}U_{A} & 0 & U_{C} & 0\\ 0 & U_{B} & 0 & > U_{D} \end{bmatrix}}_{X}\underbrace{\begin{bmatrix}S_{A} & 0 & 0 & 0\\ > 0 & S_{B} & 0 & 0\\ 0 & 0 & S_{C} & 0\\ 0 & 0 & 0 & S_{D} > \end{bmatrix}}_{S}\underbrace{\begin{bmatrix}V_{A}^{T} & 0\\ V_{B}^{T} > & 0\\ 0 & V_{C}^{T}\\ 0 & V_{D}^{T} \end{bmatrix}}_{Y^{T}}\tag{*} $$

$$ rank(M) \leq rank(A) + rank(B) + rank(C) + rank(Z) \tag{2} $$

with $Z = D- BA^{-1}C $,

We want to determine the eigenvalues of $A_{ij}$

Part One $A_{11}$

If you take $A_{11}$

$$ A_{11}^{T}A_{11} = \begin{bmatrix} 6& -5 & 0 \\ -5 & 10 & -1 \\0 & -1&2 \end{bmatrix} \tag{3}$$

$$ \det(A_{11}^{T}A_{11} - \lambda I) = \begin{vmatrix} 6- \lambda& -5 & 0 \\ -5 & 10 - \lambda & -1 \\0 & -1&2 - \lambda \end{vmatrix} \tag{4} $$

$$ det(A_{11}^{T}A_{11} -\lambda I) = (6-\lambda) \begin{vmatrix} 10- \lambda & -1 \\ -1 & 2-\lambda \end{vmatrix} +5 \begin{vmatrix} -5 & -1 \\ -0 & 2-\lambda \end{vmatrix} + 0\cdot \textrm{ stuff} \tag{5}$$

$$ \det(A_{11}^{T}A_{11}-\lambda I) = (6-\lambda)\bigg((10-\lambda)(2-\lambda) -1 \bigg) + 5 \bigg((-5)(2-\lambda) \bigg) \tag{6}$$ $$ \det(A_{11}^{T}A_{11}-\lambda I) = (6-\lambda)(\lambda^{2} -12\lambda +19) -25\lambda -50 \tag{7}$$ $$\det(A_{11}^{T}A_{11}-\lambda I) = -(\lambda^{3} -18\lambda^{2} +116\lambda -64 )\tag{8} $$

giving you three eigenvalues..which you compare against $A_{22}$

Part Two $A_{22}$

$$ A_{22}^{T}A_{22} = \begin{bmatrix} 17& 8 & 5 \\ 8 & 5 & 4 \\5 & 4&6 \end{bmatrix} \tag{9}$$

$$ \det(A_{22}^{T}A_{22} - \lambda I) = \begin{vmatrix} 17- \lambda & 8 & 5 \\ 8 & 5 -\lambda & 4 \\5 & 4&6 -\lambda \end{vmatrix} \tag{10} $$

$$ \det(A_{22}^{T}A_{22} - \lambda I) = (17-\lambda) \begin{vmatrix} 5- \lambda & 4 \\ 4 & 6-\lambda \end{vmatrix} - 8 \cdot\begin{vmatrix} 8 & 4 \\ 5 & 6-\lambda \end{vmatrix} +5 \begin{vmatrix} 8 & 5-\lambda \\ 5 & 4 \end{vmatrix} \tag{11}$$

$$ \det(A_{22}^{T}A_{22} - \lambda I) = (17-\lambda)\bigg( (5-\lambda)(6-\lambda) -16\bigg) - 8 \cdot \bigg((8)(6-\lambda) -20 \bigg)+5\bigg(40-(5-\lambda)(5)\bigg) \tag{12}$$

$$ \det(A_{22}^{T}A_{22} - \lambda I) = -\lambda^{3} +28\lambda^{2} -201\lambda +238 \\ 64\lambda -224 \\ 25\lambda + 75 \tag{13}$$

$$ \det(A_{22}^{T}A_{22} - \lambda I) = -\lambda^{3} +28\lambda^{2} -112\lambda +89 \tag{14}$$

All of these eigenvalues are distinct, so the rank is $6$. Two of them are complex actually.