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

Suppose that ${\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}}$ are eigenvectors corresponding to distinct eigenvalues ${\displaystyle \lambda _{1},\lambda _{2},\dots ,\lambda _{n}}$, respectively. Let ${\displaystyle m}$ be the largest integer for which ${\displaystyle \mathbf {v} _{1},\dots ,\mathbf {v} _{m}}$ are linearly independent. If we had ${\displaystyle m, we could write ${\displaystyle \mathbf {v} _{m+1}=\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}}$ for some ${\displaystyle c_{i}}$ not all zero. Now ${\displaystyle A\mathbf {v} _{m+1}=\lambda _{m+1}\mathbf {v} _{m+1}\implies A\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}=\sum _{i=1}^{m}c_{i}\mathbf {\lambda } _{i}\mathbf {v} _{i}=\lambda _{m+1}\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}}$, whence ${\displaystyle \sum _{i=1}^{m}c_{i}\mathbf {(} \lambda _{i}-\lambda _{m+1})\mathbf {v} _{i}=\mathbf {0} }$. As the ${\displaystyle \lambda _{i}}$ are distinct, this implies (by linear independence) that ${\displaystyle c_{1}=\cdots =c_{n}=0}$, so ${\displaystyle \mathbf {v} _{m+1}=\mathbf {0} }$, contradicting the assumption that ${\displaystyle \mathbf {v} _{m+1}}$ was an eigenvector. We conclude that ${\displaystyle m=n}$, 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 ${\textstyle \color {blue}3.}$