The Vector Space Consisting of All Traceless Diagonal Matrices

Problems and solutions in Linear Algebra

Problem 79

Let $V$ be the set of all $n \times n$ diagonal matrices whose traces are zero.
That is,

\begin{equation*}
V:=\left\{ A=\begin{bmatrix}
a_{11} & 0 & \dots & 0 \\
0 &a_{22} & \dots & 0 \\
0 & 0 & \ddots & \vdots \\
0 & 0 & \dots & a_{nn}
\end{bmatrix} \quad \middle| \quad
\begin{array}{l}
a_{11}, \dots, a_{nn} \in \C,\\
\tr(A)=0 \\
\end{array}
\right\}
\end{equation*}

Let $E_{ij}$ denote the $n \times n$ matrix whose $(i,j)$-entry is $1$ and zero elsewhere.

(a) Show that $V$ is a subspace of the vector space $M_n$ over $\C$ of all $n\times n$ matrices. (You may assume without a proof that $M_n$ is a vector space.)

(b) Show that matrices
\[E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}\] are a basis for the vector space $V$.

(c) Find the dimension of $V$.
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Hint (Subspace Criteria)

For (a), use the criteria for a subset to be a subspace.

  1. The zero vector $\mathbf{0} \in M_n$ is in $V$.
  2. For $\mathbf{u}, \mathbf{v}\in V$, the sum $\mathbf{u}+\mathbf{v}\in V$.
  3. For $\mathbf{v}\in V, c\in \C$, the scalar product $c\mathbf{v}\in V$.

Proof.

(a) $V$ is a subspace of the vector space $M_n$ over $\C$

We check the following criteria for a subset to be a subspace.

  1. The zero vector $\mathbf{0} \in M_n$ is in $V$.
  2. For $\mathbf{u}, \mathbf{v}\in V$, the sum $\mathbf{u}+\mathbf{v}\in V$.
  3. For $\mathbf{v}\in V, c\in \C$, the scalar product $c\mathbf{v}\in V$.

For (1), note that the zero vector $\mathbf{0}\in M_n$ is the $n \times n$ zero matrix. Thus it is clearly in $V$.


To check (2), take $\mathbf{u}, \mathbf{v}\in V$. Then $\mathbf{u}, \mathbf{v}$ are diagonal matrices and $\tr(\mathbf{u})=0$, $\tr(\mathbf{v})=0$.
Thus the sum $\mathbf{u}+\mathbf{v}$ is a diagonal matrix and the trace is
\[\tr(\mathbf{u}+\mathbf{v})=\tr(\mathbf{u})+\tr(\mathbf{v})=0+0=0.\] Hence the sum $\mathbf{u}+\mathbf{v} \in V$.


Finally, to prove (3) we take $\mathbf{v}\in V, c\in \C$.
Then $\mathbf{v}$ is a diagonal matrix and $\tr(\mathbf{v})=0$.
Thus $c\mathbf{v}$ is also a diagonal matrix and the trace is
\[\tr(c\mathbf{v})=c\tr(\mathbf{v})=c\cdot 0=0.\] Hence $c\mathbf{v} \in V$.
Therefore the subspace criteria (1)-(3) hold. Thus $V$ is a subspace of the vector space $M_n$.

(b) A basis for the vector space $V$.

Note that for each $1 \leq i \leq n-1$, the matrix $E_{ii}-E_{i+1\, i+1}$ is a diagonal matrix and the trace is
\[\tr(E_{ii}-E_{i+1\, i+1})=\tr(E_{ii})-\tr(E_{i+1\,i+1})=1-1=0.\] Hence the matrix $E_{ii}-E_{i+1\, i+1} \in V$.

We show that the matrices $E_{ii}-E_{i+1\, i+1}$ span $V$.
Let $A=(a_{ij})\in V$. We want to solve the following linear combination.
\begin{align*}
A=c_1(E_{11}-E_{22})+c_2(E_{22}-E_{33})+\cdots+c_{n-1}( E_{n-1\, n-1}-E_{nn}).
\end{align*}
Rearranging this, we obtain
\begin{align*}
A=c_1E_{11}+(c_2-c_1)E_{22}+\cdots+(c_{n-1}-c_{n-2}) E_{n-1\, n-1}-c_{n-1}E_{nn}.
\end{align*}


Namely, in the matrix form we have
\[\begin{bmatrix}
a_{11} & 0 & \dots & 0 \\
0 &a_{22} & \dots & 0 \\
\vdots & \ddots & \ddots & \ddots \\
0 & 0 & \dots & a_{nn}
\end{bmatrix}
=
\begin{bmatrix}
c_1 & 0 & \dots & \dots &0 \\
0 & c_2-c_1 & \dots & \dots & 0 \\
\vdots & \ddots & \ddots & \ddots & \vdots \\
0 & \dots & 0 & c_{n-1}-c_{n-2} & 0 \\
0 & \dots & \dots & 0 & -c_{n-1}
\end{bmatrix}.\] Comparing entries, we have the system of $n$ equations in $n-1$ unknowns $c_1,\dots, c_{n-1}$
\begin{align*}
a_{11}&=c_1\\
a_{22}&=c_2-c_1\\
a_{33}&=c_3-c_2\\
\vdots\\
a_{n-1 n-1}&=c_{n-1}-c_{n-2}\\
a_{nn}&=-c_{n-1}.
\end{align*}


We solve this system as follows.
The matrix form of this system is
\[\begin{bmatrix}
1 & & & & \\
-1 & 1 & & & \\
& -1 & 1 & & \\
& & \ddots & \ddots & \\
& & & -1& 1\\
& & & & -1
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2 \\
\vdots \\
c_{n-1}
\end{bmatrix}=\begin{bmatrix}
a_{11} \\
a_{22} \\
\vdots \\
a_{nn}
\end{bmatrix}.\] (The empty entries are all zero.)


The augmented matrix is
\[\left[\begin{array}{rrrrr|r}
1 & & & & & a_{11}\\
-1 & 1 & & & &a_{22} \\
& -1 & 1 & & &a_{33}\\
& & \ddots & \ddots & & \vdots \\
& & & -1& 1 &a_{n-1 n-1}\\
& & & & -1 &a_{n n}
\end{array} \right].\] Then we reduce this matrix as follows.
Add the first row to the second row, then add the second row to the third row.
Repeating this process we get
\[\left[\begin{array}{rrrrr|r}
1 & & & & & a_{11}\\
& 1 & & & &a_{11}+a_{22} \\
& & 1 & & &a_{11}+a_{22}+a_{33}\\
& & \ddots & \ddots & & \vdots \\
& & & & 1 &a_{11}+a_{22}+\dots +a_{n-1 n-1}\\
& & & & 0 &a_{11}+a_{22}+\dots +a_{n n}
\end{array} \right].\]


Note that since $A \in V$, $\tr(A)=a_{11}+a_{22}+\cdots a_{nn}=0$, thus the right-bottom entry ($(n,n)$-entry) of the augmented matrix is $0$, hence the system is consistent and it has the unique solution
\begin{align*}
c_{1} &= a_{11} \\
c_{2} &=a_{11}+a_{22}\\
c_3 &=a_{11}+a_{22}+a_{33}\\
\vdots \\
c_{n-1}&=a_{11}+a_{22}+\cdots+a_{n-1 n-1}.\tag{*}
\end{align*}
Therefore the matrices $E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}$ span $V$.


To show that these matrices are linearly independent, we can reuse the above computation.
Suppose we have a linear combination
\[\mathbf{0}=c_1(E_{11}-E_{22})+c_2(E_{22}-E_{33})+\cdots+c_{n-1}( E_{n-1\, n-1}-E_{nn}).\] Here $\mathbf{0}$ is the $n \times n$ zero matrix.
Since $a_{ii}=0$ in the above computation, we see from (*) that $c_1=c_2=\dots=c_{n-1}=0$, hence the matrices $E_{ii}-E_{i+1 i+1}$ are linearly independent.
Therefore those matrices are a basis of $V$.

(c) The dimension of $V$.

From part (b), the $n-1$ matrices
\[E_{11}-E_{22}, \, E_{22}-E_{33}, \, \dots,\, E_{n-1\, n-1}-E_{nn}\] are a basis of $V$. Thus the dimension of $V$ is $n-1$.


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