The number of linearly independent eigen vectors of \[\left[ {\begin{array}{*{20}{c}} 2&1 \\ 0&2 \end{array}} \right]\] is A. 0 B. 1 C. 2 D. infinite

0
1
2
infinite

The correct answer is $\boxed{\text{C}}$.

The characteristic polynomial of the matrix $A$ is $p(x) = x^2 – 4x + 4 = (x – 2)^2$. This means that the eigenvalues of $A$ are $2$ and $2$, with multiplicities $2$ and $1$, respectively.

A vector $v$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ if $Av = \lambda v$. In this case, the eigenvectors corresponding to the eigenvalue $2$ are $(1, 0)$ and $(0, 1)$. These two vectors are linearly independent, so there are $2$ linearly independent eigenvectors of $A$.

Option A is incorrect because the matrix $A$ has two distinct eigenvalues. Option B is incorrect because the matrix $A$ has two linearly independent eigenvectors. Option D is incorrect because the matrix $A$ is not a diagonalizable matrix.