Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (c)/Solution 1

In fact, eigenvectors corresponding to distinct eigenvalues are linearly independent. Let’s prove it.

Suppose that $\mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}$ are eigenvectors corresponding to distinct eigenvalues $\lambda _{1},\lambda _{2},\dots ,\lambda _{n}$, respectively. Let $m$ be the largest integer for which $\mathbf {v} _{1},\dots ,\mathbf {v} _{m}$ are linearly independent. If we had $m<n$, we could write $\mathbf {v} _{m+1}=\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}$ for some $c_{i}$ not all zero. Now $A\mathbf {v} _{m+1}=\lambda _{m+1}\mathbf {v} _{m+1}\implies A\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}=\sum _{i=1}^{m}c_{i}\mathbf {\lambda } _{i}\mathbf {v} _{i}=\lambda _{m+1}\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}$, whence $\sum _{i=1}^{m}c_{i}\mathbf {(} \lambda _{i}-\lambda _{m+1})\mathbf {v} _{i}=\mathbf {0}$. As the $\lambda _{i}$ are distinct, this implies (by linear independence) that $c_{1}=\cdots =c_{n}=0$, so $\mathbf {v} _{m+1}=\mathbf {0}$, contradicting the assumption that $\mathbf {v} _{m+1}$ was an eigenvector. We conclude that $m=n$, so all the eigenvectors are linearly independent.

Each of the 3 eigenvalues found in part (b) has a corresponding eigenvector, and by the above, they are linearly independent. (Clearly, there cannot be more linearly independent eigenvectors, since the dimension of the matrix is 3.) Thus, the answer is ${\textstyle \color {blue}3.}$