True, false questions about matrices

[lukaszh]

Hello,
could you help me with some true false questions about matrices?

First
If $$\mathbf{A}=\mathbf{B}-\mathrm{i}\mathbf{C}$$ is hermitian matrix $$\mathbf{B},\mathbf{C}$$ are real, then $$\mathbf{B},\mathbf{C}$$ are anti-symmetric matrices. True? False?

My solution
If $$\mathbf{A}$$ is hermitian, then $$\mathbf{A}^{\mathrm{H}}=\mathbf{A}$$ so $$(\mathbf{B}-\mathrm{i}\mathbf{C})^{\mathrm{H}}=\mathbf{B}^{\mathrm{H}}-(\mathrm{i}\mathbf{C})^{\mathrm{H}}=\mathbf{B}^{\mathrm{T}}+i\mathbf{C}^{\mathrm{T}}$$. It implies the fact $$\mathbf{B}^{\mathrm{T}}+i\mathbf{C}^{\mathrm{T}}=\mathbf{B}-\mathrm{i}\mathbf{C}$$. $$\mathbf{B}$$ is symmetric and $$\mathbf{C}$$ is anti-symmetric. FALSE

Second
If A is diagonalizable and its eigenvalues are $$\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$$, then $$\prod_{k=1}^{n}(x-\lambda_k)=0$$ has n different solutions

My solution
Matrix is diagonalizable, then $$\lambda_i\ne\lambda_j$$ for $$i\ne j$$. So polynomial has n different soln's. TRUE


Thank you very much for your help...

 

[HallsofIvy]

You have a much bigger problem that whether this statement is true of false! What makes you think that if a matrix is diagonalizable, then $\lambda_i\ne\lambda_j$?

The identity matrix is diagonalizable because it is diagonal. What are its eigenvalues?

(An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. It doesn't matter what the eigenvalues are.)