# Question 2

(i) We have:

T(x,y,z,t) = \begin{pmatrix}x - y + z + t \\ x + 2y - z + t \\ 3y - 2z\end{pmatrix}

So T(x,y,z,t) = 0 iff x - y + z + t = 0, x + 2y - z + t = 0 and 3y - 2z = 0. Note that subtracting the first two equations gives the third, so the third is in fact redundant. The solution set can therefore be determined only by x - y + z + t = 0 and x + 2y - z + t = 0. This is undetermined and we are free to set two variables arbitrary, say x = \lambda and y = \mu. Then \displaystyle z = \frac 3 2 \mu and \displaystyle t = y - x - z = \mu - \lambda - \frac 3 2 \mu = -\frac 1 2 \mu - \lambda. So each vector in X can be written:

\displaystyle \begin{pmatrix}\lambda \\ \mu \\ \frac 3 2 \mu \\ -\frac 1 2 \mu - \lambda\end{pmatrix} = \lambda \begin{pmatrix}1 \\ 0 \\ 0 \\ -1\end{pmatrix} + \mu \begin{pmatrix}0 \\ 1 \\ \frac 3 2 \\ -\frac 1 2\end{pmatrix}

So:

\mathrm{ker}(T) = \mathrm{span} \left\{\begin{pmatrix}1 \\ 0 \\ 0 \\ -1\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ \frac 3 2 \\ -\frac 1 2\end{pmatrix}\right\}

(note that these are linearly independent so it in fact forms a basis)

For the image of T, we are interested in:

\mathrm{span} \left\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}, \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix}\right\}

Note that \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} is not a multiple of \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}. It remains to investigate whether \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix} is a linear combination of these two. We investigate solutions to:

\begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix} = \lambda \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} + \mu \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}

The last component requires \mu = -\dfrac 2 3 and the first requires \lambda = \dfrac 1 3. This is consistent with the second component, so \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix} can be written as a linear combination of the other two vectors. So:

\mathrm{im}(T) = \mathrm{span} \left\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}, \begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix}\right\} = \mathrm{span} \left\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 2 \\ 3\end{pmatrix}\right\}

So we have \mathrm{rank}(T) = 2 and \mathrm{Nullity}(T) = 2.

(ii)

Note that we are free to determine x_{ii} for 1 \le i \le n - 1 arbitrarily, then x_{nn} is fixed and dependent on the other x_{ii} s. The other elements of the matrix can be decided arbitrarily. Then a basis for \mathrm{ker}(T) is:

\left\{D_i \mid 1 \le i \le n - 1\right\} \cup \left\{T_{ij} \mid 1 \le i, j \le n, \, i \ne j\right\}

where:

• D_i has all elements zero except a 1 in the (i, i) position and -1 in the (n,n) position.
• T_{ij} has all elements zero except a 1 in the (i, j) position.

(noting linear independence here)

Note that since these sets are disjoint the size of the union is:

\mathrm{Nullity}(T) = |\left\{D_i \mid 1 \le i \le n - 1\right\}| + |\left\{T_{ij} \mid 1 \le i, j \le n, \, i \ne j\right\}|

The former set clearly has size n - 1. For the latter set, note that there are n^2 possible ways to pair the i, j s ignoring the requirement that i \ne j. There are n pairs with i = j, so the second set has size n^2 - n.

Hence, we have \mathrm{Nullity}(T) = n^2 - 1.

We can show T to be surjective. Let t \in \mathbb R. Then let X = (x_{ij}) be the matrix with x_{11} = t and x_{ij} = 0 for (i,j) \ne (1,1). Then \mathrm{tr}(X) = t. So T is surjective, and \mathrm{im}(T) = \mathbb R. This has dimension 1, so \mathrm{rank}(T) = 1.