# Matrix Trace

Oct 7, 2020
Sep 15, 2021 15:55 UTC
Such a simple concept with so many properties and applications!

The trace of a square matrix $\boldsymbol{A}: n \times n$ is the sum of the diagonals:

$$tr(\boldsymbol{A}) = \sum_{i=1}^n a_{ii}$$

The trace is related to many important features of matrices, and we’re going to discuss some basic properties first.

## Properties

• Trace is commutative, i.e. $tr(\boldsymbol{AB}) = tr(\boldsymbol{BA})$. Note that if $\boldsymbol{A}: m \times n$ and $\boldsymbol{B}: n \times m$, $\boldsymbol{AB}$ would be $m \times m$ and $\boldsymbol{BA}$ would be $n \times n$.

\begin{aligned} LHS &= \sum_{i=1}^m (\boldsymbol{AB})_{ii} \\ &= \sum_{i=1}^m \left( \sum_{i=1}^n a_{ij}b_{ji} \right) \\ &= \sum_{j=1}^n \sum_{i=1}^m b_{ji}a_{ij} \\ &= \sum_{j=1}^n (\boldsymbol{BA})_{jj} = tr(\boldsymbol{BA}) \end{aligned}

For more than two matrices, note that $tr(\boldsymbol{ABC}) \neq tr(\boldsymbol{CBA})$. Instead,

$$tr(\boldsymbol{ABC}) = tr(\boldsymbol{A}(\boldsymbol{BC})) = tr(\boldsymbol{BCA})$$

Similarly, we have

$$tr(\boldsymbol{ABCDE}) = tr(\boldsymbol{CDEAB}) = tr(\boldsymbol{EABCD}) = \cdots$$

• $tr(\boldsymbol{A}^\prime \boldsymbol{A}) = tr(\boldsymbol{AA}^\prime) \geq 0$.

This is equal to the sum of all entries squared.

• $tr(\boldsymbol{A}^\prime\boldsymbol{A}) = 0$ is equivalent to $\boldsymbol{A} = 0$.

Showing $\Leftarrow$ is trivial. For $\Rightarrow$, the trace being zero means that the sum of all entries squared of $\boldsymbol{A}$ is zero, thus $a_{ij} = 0$ for all $i, j$.

We can also use this fact to show that $tr(\boldsymbol{A}^\prime \boldsymbol{A}) = 0 \Leftrightarrow \boldsymbol{A}^\prime\boldsymbol{A} = 0$.

• $\boldsymbol{A}^\prime \boldsymbol{AB} = \boldsymbol{A}^\prime \boldsymbol{AC} \Leftrightarrow \boldsymbol{AB} = \boldsymbol{AC}$.

This one is related to future statistical analyses (e.g. linear regression). Again $\Leftarrow$ is obvious. To prove $\Rightarrow$, we will show $\boldsymbol{AB} = \boldsymbol{AC}$ by seeing that $tr((\boldsymbol{AB} - \boldsymbol{AC})^\prime (\boldsymbol{AB} - \boldsymbol{AC})) = 0$.

We start with $\boldsymbol{A}^\prime \boldsymbol{AB} - \boldsymbol{A}^\prime \boldsymbol{AC} = 0$.

$$\begin{gathered} \boldsymbol{A}^\prime (\boldsymbol{AB} - \boldsymbol{AC}) = 0 \\ (\boldsymbol{B}^\prime - \boldsymbol{C}^\prime) \boldsymbol{A}^\prime (\boldsymbol{AB} - \boldsymbol{AC}) = 0 \\ (\boldsymbol{AB} - \boldsymbol{AC})^\prime (\boldsymbol{AB} - \boldsymbol{AC}) = 0 \\ \boldsymbol{AB} - \boldsymbol{AC} = 0 \end{gathered}$$

We will use this fact to show that $\mathcal{C}(\boldsymbol{A}) = \mathcal{C}(\boldsymbol{AA}^\prime)$.

### Recap on column spaces

Say $\boldsymbol{A}$ is an $m \times n$ matrix. If vector $\boldsymbol{y} \in \mathcal{C}(\boldsymbol{A})$, it means $\boldsymbol{y} = \boldsymbol{Ax}$ for some $\boldsymbol{x}$. Similarly if $\boldsymbol{y} \in \mathcal{C}(\boldsymbol{A})^\perp$, then $\boldsymbol{y} \perp \mathcal{C}(\boldsymbol{A})$. Since $\boldsymbol{y}$ is orthogonal to each column of $\boldsymbol{A}$, we have

$$\boldsymbol{y}^\prime \boldsymbol{a}_j = 0 \quad \forall j = 1, \cdots, n \Rightarrow \boldsymbol{y}^\prime \boldsymbol{A} = \boldsymbol{0}^\prime$$

Similarly, if $\boldsymbol{y}^\prime \in \mathcal{R}(\boldsymbol{A})$, then $\boldsymbol{y}^\prime = \boldsymbol{x}^\prime \boldsymbol{A}$ for some $\boldsymbol{x}$, and

$$\boldsymbol{y}^\prime \in \underbrace{\mathcal{R}(\boldsymbol{A})^\perp}_{=N(\boldsymbol{A})} \Leftrightarrow \boldsymbol{Ay} = \boldsymbol{0}_{m \times 1}$$

• Suppose $\mathcal{W}, \mathcal{V}$ are subspaces in $\mathbb{R}^n$, and $\mathcal{W} \subset \mathcal{V}$. Then, $\mathcal{V}^\perp \subset \mathcal{W}^\perp$.

We can express the subspaces as spans:

$$\begin{gathered} \mathcal{W} = \mathcal{L}(\boldsymbol{u}_1, \cdots, \boldsymbol{u}_k) \\ \mathcal{V} = \mathcal{L}(\boldsymbol{u}_1, \cdots, \boldsymbol{u}_k, \boldsymbol{u}_{k+1}, \cdots, \boldsymbol{u}_m) \end{gathered}$$

For any $\boldsymbol{y} \in \mathcal{V}^\perp$, we have

$$\boldsymbol{y} \perp \mathcal{V} \Leftrightarrow \boldsymbol{y} \perp \boldsymbol{u}_i, i = 1, \cdots, m$$

Which means $\boldsymbol{y} \perp \boldsymbol{u}_i, i = 1, \cdots, k$, thus $\boldsymbol{y} \perp \mathcal{W}$ and $\boldsymbol{y} \in \mathcal{W}^\perp$.

### Proof

We’re now ready to prove $\mathcal{C}(\boldsymbol{A}^\prime) = \mathcal{C}(\boldsymbol{A}^\prime \boldsymbol{A})$. Usually is not that easy to prove set equality directly. Often we prove the LHS is a subset of the RHS and vice versa.

We first show that $\mathcal{C}(\boldsymbol{A}^\prime \boldsymbol{A}) \subset \mathcal{C}(\boldsymbol{A}^\prime)$. This is obvious because $\mathcal{C}(\boldsymbol{AB}) \subset \mathcal{C}(\boldsymbol{A})$.

To show that $\mathcal{C}(\boldsymbol{A}^\prime) \subset \mathcal{C}(\boldsymbol{A}^\prime \boldsymbol{A})$, we will show $\mathcal{C}(\boldsymbol{A}^\prime \boldsymbol{A})^\perp \subset \mathcal{C}(\boldsymbol{A}^\prime)^\perp$ instead. For all $\boldsymbol{y} \in \mathcal{C}(\boldsymbol{A}^\prime \boldsymbol{A})^\perp$,

$$\begin{gathered} \quad \boldsymbol{y}^\prime \boldsymbol{A}^\prime \boldsymbol{A} = \boldsymbol{0}^\prime \xrightarrow{transpose} \boldsymbol{A}^\prime \boldsymbol{Ay} = \boldsymbol{0}_{n \times 1} \\ \Rightarrow \boldsymbol{Ay} = \boldsymbol{0}_{m \times 1} \xrightarrow{transpose} \boldsymbol{y}^\prime \boldsymbol{A} = \boldsymbol{0}^\prime \\ \boldsymbol{y} \in \mathcal{C}(\boldsymbol{A}^\prime)^\perp \end{gathered}$$

Similarly, we can show that

$$\begin{gathered} \mathcal{C}(\boldsymbol{A}) = \mathcal{C}(\boldsymbol{AA}^\prime) \\ \mathcal{R}(\boldsymbol{A}) = \mathcal{R}(\boldsymbol{A}^\prime \boldsymbol{A}) \\ r(\boldsymbol{A}) = r(\boldsymbol{AA}^\prime) = r(\boldsymbol{A}^\prime\boldsymbol{A}) \end{gathered}$$

## One-way ANOVA example

Suppose matrix $\boldsymbol{A}$ is given by

$$\boldsymbol{A} = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix}$$

We can easily see that the second the third columns are linearly independent, but the first column is the sum of the other two columns, so $r(\boldsymbol{A}) = 2$. The column space of $\boldsymbol{A}$ is a subspace of $\mathbb{R}^4$. Suppose the data we collected is

$$\boldsymbol{y} = \begin{pmatrix} 2 \\ 1 \\ 3 \\ 5 \end{pmatrix}$$

In ANOVA, we’re going to need two more matrices:

$$\boldsymbol{A}^\prime \boldsymbol{A} = \begin{pmatrix} 4 & 2 & 2 \\ 2 & 2 & 0 \\ 2 & 0 & 2 \end{pmatrix}, \quad \boldsymbol{A}^\prime \boldsymbol{y} = \begin{pmatrix} 11 \\ 3 \\ 8 \end{pmatrix}$$

and we need to solve for $\boldsymbol{\beta} \in \mathbb{R}^3$:

$$\boldsymbol{A}^\prime \boldsymbol{A\beta} = \boldsymbol{A}^\prime \boldsymbol{y}$$

We’re interested in how to solve this, if the solution is unique, or does a solution even exist. The next chapter is dedicated to linear systems.

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