# When eigenvectors for a matrix form a basis

It is well known that if n by n matrix A has n distinct eigenvalues, the eigenvectors form a basis.

Also, if A is symmetric, the same result holds.

Consider

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

This matrix has single eigenvalue $\lambda=1$, and is not symmetric.

But, the eigenvectors corresponding to $\lambda=1$, ($v_1 =\left[ {\begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array}}\right]$ , $v_2 =\left[ {\begin{array}{c} 0 \\ 1/2 \\ 0 \\ \end{array}}\right]$ , $v_3 =\left[ {\begin{array}{c} 0 \\ -3/8 \\ 1/4 \\ \end{array}}\right]$ ) form a basis.

What sufficient conditions offer the above result?

• Non of $\;v_2,\,v_3\;$ is an eigenvector of $\;A\;$ wrt $\;\lambda=1\;$ ...In fact, your $\;A\;$ has only one linearly independent eigenvector wrt to its unique eigenvalue, which can be $\;v_1\;$ . Commented May 9, 2017 at 9:10
• This is the definition of diagonalisable. There are many things known about diagonalisability; read your textbook, this question is not asking about anything specific. I am voting to close as "not clear what you are asking". Commented May 9, 2017 at 11:51

A square matrix is diagonalizable if and only if there exists a basis of eigenvectors. That is, $A$ is diagonalizable if there exists an invertible matrix $P$ such that $P^{-1}AP=D$ where $D$ is a diagonal matrix.
Let $A$ be the matrix of a linear transformation $\alpha: V\to V$, where $V$ is an $n$-dimensional vector space over the field $\mathbb{F}$.
$V$ has a basis consisting of eigenvectors of $\alpha$ (or $A$ if you prefer) if, and only if, the minimal polynomial $m_\alpha(X)$ of $\alpha$ (or of $A$ if you prefer) is a product of distinct linear factors in $\mathbb{F}[X]$.