Linear Algebra: Companion notes

The rank of \(A\) is \(1\text{.}\) By rank-nullity theorem, nullity of \(A\) is \(2\text{.}\) Hence, \(0\) is an eigenvalue of \(A\text{.}\) In fact we have \(\chi_A=t^2(t-3)\text{.}\) Therefore, the algebraic multiplicity of \(0\) and \(3\) is respectively, \(2\) and \(1\text{.}\)

The geometric multiplicity of \(0\text{,}\) \(\dim_\C(\ker A)\) is \(2\) and the geometric multiplicity of \(1\text{,}\) \(\dim_\C(\ker(A-3I_3))\) is \(1\text{.}\) Hence the number of Jordan block corresponding to \(3\) is \(1\) and the number of Jordan block corresponding to \(0\) is \(2\text{.}\)

In this case we get a basis consisting of eigenvectors. So the matrix is diagonalizable. Anyhow, the minimal polynomial of \(A\) is \(t(t-3)\text{.}\) Hence, the largest Jordan block corresponding to \(0\) and \(3\) is of size \(1\text{.}\)

We have \(\chi_A=(t-1)^2\in\C[t]\text{.}\) The algebraic multiplicity of \(\lambda=1\) is \(2\text{.}\) We have \(A-I_2=\begin{pmatrix}0\amp 1\\0\amp 0\end{pmatrix}\text{.}\) Hence the geometric multiplicity of \(\lambda=1\) is \(1\text{.}\) Thus there is one Jordan block, i.e., in this case, \(A\) is similar to a Jordan block. Since we can not have a basis consisting of an eigenvectors, the Jordan form must be \(\begin{pmatrix}1\amp 0\\1\amp 1\end{pmatrix}\text{.}\)

Anyway, let us compute the minimal polynomial. Since \(A-I_2\neq 0\text{,}\) the minimal polynomial is \((t-1)^2\text{.}\) Hence the largest Jordan block corresponding to \(\lambda=1\) has size \(2\text{.}\)

The rank of \(A\) is \(1\text{,}\) hence the nullity of \(A\) is \(2\text{.}\) Therefore, \(0\) is an eigenvalue. Furthermore the algebraic and geometric multiplicity of \(\lambda=0\) is \(2\text{.}\) Let \(\lambda_1=\lambda_2=0\text{.}\) We can find \(\lambda_3\) using trace. Indeed, \(-2=tr(A)=\lambda_1+\lambda_2+\lambda_3=\lambda_3\text{.}\)

The geometric multiplicity of \(0\) is \(2\text{.}\) Hence there are two Jordan blocks corresponding to \(0\text{.}\) The geometric multiplicity of \(-2\) is \(1\) hence there is only one Jordan block corresponding to this eigenvalue. Thus we can see that \(A\) is diagonalizable and thus, \(A\) is similar to \({\rm diag}(0,0,-2)\text{.}\)

Anyway let us calculate the minimal polynomial. It is \(t(t+2)\text{.}\) Therefore the largest Jordan block corresponding to eigenvalues \(0\) and \(-2\) is of the size \(1\text{.}\)

All eigenvalues of \(A\) are equal to \(2\text{.}\) The algebraic multiplicity of \(2\) is \(3\text{.}\)

The geometric multiplicity of \(2\) is the dimension of the kernel of \(A-2I_2=\begin{pmatrix}3\amp 4\amp 1\\-1\amp -2\amp 0\\-3\amp -4\amp -1\end{pmatrix}\text{.}\) The row-reduced echelon form of this matrix isTherefore, the geometric multiplicity of \(\lambda=2\) is \(1\text{.}\) Hence the number of Jordan blocks corresponding to \(\lambda=2\) is \(1\text{.}\) Hence, \(A\) must be similar to

Note that \((A-2I_2)^2\neq 0\) hence the minimal polynomial of \(A\) must be \((A-2I_2)^3\text{.}\) Thus the largest Jordan block corresponding to \(\lambda=2\) is of the size \(3\text{.}\)

Eigenvalues of \(A\) are \(1,1,-1,-1\text{.}\) Thus the algebraic multiplicity of both \(1\) and \(-1\) is \(2\text{.}\)

To find geometric multiplicity of \(1\) and \(-1\text{.}\) We haveThus, nullity of \(A-I_4\text{,}\) i.e., the geometric multiplicity of \(\lambda=1\) is \(2\text{.}\) Hence, the number of Jordan blocks corresponding to \(\lambda=1\) is \(2\text{.}\) On the other hand, computing the row-reduced echelon form of \(A+I_4\text{,}\) we get that the nullity of \(A+I_4\) is \(1\text{.}\) Hence the geometric multiplicity of \(\lambda=-1\) is \(1\text{.}\) Therefore, the number of Jordan block corresponding to \(\lambda=-1\) is \(1\text{.}\) Thus \(A\) is similar to

All eigenvalues of \(A\) are equal to \(1\text{.}\) The algebraic multiplicity of \(\lambda=1\) is \(4\text{.}\)

ConsiderHence the nullity of \(A-I_4\) is \(2\text{.}\) Thus there are two Jordan blocks corresponding to \(\lambda=1\text{.}\)

We have \((A-I_4)^2=0\) and \(A-I_4\neq 0\text{.}\) Hence the minimal polynomial of \(A\) is \((t-1)^2\text{.}\) Therefore, size of the largest Jordan block corresponding to \(\lambda=1\) is \(2\text{.}\)