The eigenvectors for the matrix comprise the eigenspace of the matrix when the zero vector is also included (Text, Definition EM, p. 377). This is equivalent to the null space of the matrix at the heart of the characteristic polynomial (Textbook, Theorem EMNS, p. 378), or \(\mathcal{E}_A(\lambda)=\mathcal{N}(A - \lambda I)\).
First let \(\lambda = 1\): \[ \begin{aligned} A - I_3 &= \begin{bmatrix} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 2 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 0 & 0\\ 0 & \fbox{1} & 0\\ 0 & 0 & \fbox{1} \end{bmatrix}\\ \mathcal{E}_A(1)&=\mathcal{N}(A - I_3) = \left\langle \left\{ \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\right\}\right\rangle \end{aligned} \] Note that this is already a unit vector.
Next let \(\lambda = 4\): \[ \begin{aligned} A - 4I_3 &= \begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2 \end{bmatrix} \Rightarrow \begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} -3 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} \fbox{1} & -\frac{2}{3} & 0\\ 0 & 0 & \fbox{1}\\ 0 & 0 & 0 \end{bmatrix}\\ \mathcal{E}_A(4)&=\mathcal{N}(A - 4I_3) = \left\langle \left\{ \begin{bmatrix} \frac{2}{3}\\ 1\\ 0 \end{bmatrix}\right\}\right\rangle \end{aligned} \] The norm of this vector is \(\sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}\) so the corresponding unit vector would be: \[ \left\langle \left\{ \begin{bmatrix} \frac{2}{\sqrt{13}}\\ \frac{3}{\sqrt{13}}\\ 0 \end{bmatrix}\right\}\right\rangle \]
Lastly let \(\lambda = 6\): \[ \begin{aligned} A - 6I_3 &= \begin{bmatrix} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} -5 & 0 & 8\\ 0 & -2 & 5\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} \fbox{1} & 0 & -1.6\\ 0 & \fbox{1} & -2.5\\ 0 & 0 & 0 \end{bmatrix}\\ \mathcal{E}_A(6)&=\mathcal{N}(A - 6I_3) = \left\langle \left\{ \begin{bmatrix} 1.6\\ 2.5\\ 1 \end{bmatrix}\right\}\right\rangle \end{aligned} \]
The norm of this vector is \(\frac{\sqrt{981}}{10}\) so the corresponding unit vector would be: \[ \left\langle \left\{ \begin{bmatrix} \frac{16}{\sqrt{981}}\\ \frac{25}{\sqrt{981}}\\ \frac{10}{\sqrt{981}} \end{bmatrix}\right\}\right\rangle \]
The eigenvectors for the matrix comprise the eigenspace of the matrix when the zero vector is also included (Text, Definition EM, p. 377). This is equivalent to the null space of the matrix at the heart of the characteristic polynomial (Textbook, Theorem EMNS, p. 378), or \(\mathcal{E}_A(\lambda)=\mathcal{N}(A - \lambda I)\).
First let \(\lambda = 1\): \[ \begin{aligned} A - I_3 &= \begin{bmatrix} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 2 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 0 & 0 & 0\\ 0 & \fbox{1} & 0\\ 0 & 0 & \fbox{1} \end{bmatrix}\\ \mathcal{E}_A(1)&=\mathcal{N}(A - I_3) = \left\langle \left\{ \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\right\}\right\rangle \end{aligned} \] Note that this is already a unit vector.
Next let \(\lambda = 4\): \[ \begin{aligned} A - 4I_3 &= \begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2 \end{bmatrix} \Rightarrow \begin{bmatrix} -3 & 2 & 3\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} -3 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} \fbox{1} & -\frac{2}{3} & 0\\ 0 & 0 & \fbox{1}\\ 0 & 0 & 0 \end{bmatrix}\\ \mathcal{E}_A(4)&=\mathcal{N}(A - 4I_3) = \left\langle \left\{ \begin{bmatrix} \frac{2}{3}\\ 1\\ 0 \end{bmatrix}\right\}\right\rangle \end{aligned} \] The norm of this vector is \(\sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}\) so the corresponding unit vector would be: \[ \left\langle \left\{ \begin{bmatrix} \frac{2}{\sqrt{13}}\\ \frac{3}{\sqrt{13}}\\ 0 \end{bmatrix}\right\}\right\rangle \]
Lastly let \(\lambda = 6\): \[ \begin{aligned} A - 6I_3 &= \begin{bmatrix} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} -5 & 0 & 8\\ 0 & -2 & 5\\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \begin{bmatrix} \fbox{1} & 0 & -1.6\\ 0 & \fbox{1} & -2.5\\ 0 & 0 & 0 \end{bmatrix}\\ \mathcal{E}_A(6)&=\mathcal{N}(A - 6I_3) = \left\langle \left\{ \begin{bmatrix} 1.6\\ 2.5\\ 1 \end{bmatrix}\right\}\right\rangle \end{aligned} \]
The norm of this vector is \(\frac{\sqrt{981}}{10}\) so the corresponding unit vector would be: \[ \left\langle \left\{ \begin{bmatrix} \frac{16}{\sqrt{981}}\\ \frac{25}{\sqrt{981}}\\ \frac{10}{\sqrt{981}} \end{bmatrix}\right\}\right\rangle \]