Prove similarity of matrix to Jordan Canonical Form

Question

I have a problem that I am unsure how to solve. How can I prove that a square matrix is similar to its Jordan canonical form. Specifically, how would I be able to show that a 2 by 2 matrix with first row "a", "b" and second row "c", "d", is similar to either the matrix with first row "lambda1", "0" and second row "0", "lambda 2" or the matrix with first row "lambda", "1" and second row "0", "lambda"?

Thanks a lot in advance!

Answer 1

For a $2\times2$ matrix $A$, it's rather simple:

• Either there are two eigenvalues $\lambda_1,\lambda_2$, to which correspond two (linearly independent) eigenvectors $v_1, v_2$. These make up a basis of $K^2$ ($K$ is the base field) and in this basis, by definition of eigenvectors, the matrix takes the form: $$\begin{pmatrix} \lambda_1&0\\0&\lambda_2\end{pmatrix}. $$
• Or there is only one eigenvalue $\lambda$, and there two possibilities: the eigenspace has dimension two, and in basis of eigenvectors, the matrix is equal to $\lambda$.
• Or there is only one eigenvalue $\lambda$ and the eigenspace has dimension $1$, so the characteristic polynomial of $A$ is $(t-\lambda)^2$.
  In this case, $\ker(A-\lambda I)\varsubsetneq\ker(A-\lambda I)^2=K^2$, so take any vector $v_2\not\in\ker(A-\lambda I)$, set $v_1=Av_2-\lambda v_2$ is an eigenvector. Indeed $$ (A-\lambda I)(v_1)=(A-\lambda I)^2(v_2)=0, $$ so $Av_1=\lambda v_1,\; Av_2=\lambda v_2+v_1$, and in the basis $(v_1,v_2)$ the matrix $A$ takes the form $$\begin{pmatrix} \lambda & 1 \\ 0&\lambda\end{pmatrix}.$$