# A Linear Algebra Problem on Twitter

Problem: Let $A$ and $B$ be $n\times n$ matrices such that their sum $A+B$ is invertible. Then show that $$A(A+B)^{-1}B=B(A+B)^{-1}A$$ (Hat tip: Sam Walters)

Solution. \begin{aligned}I&=(A+B)(A+B)^{-1}\\&=A(A+B)^{-1}+B(A+B)^{-1}\end{aligned}\label{eq:matrix} Multiply \eqref{eq:matrix} by $B$ from the right $$\label{eq:matrix2}B=A(A+B)^{-1}B+B(A+B)^{-1}B$$ Also multiply \eqref{eq:matrix} by $A$ from the left $$\label{eq:matrix3}A=A(A+B)^{-1}A+B(A+B)^{-1}A$$ Subtract \eqref{eq:matrix3} from \eqref{eq:matrix2}. $$\label{eq:matrix4}B-A=A(A+B)^{-1}B-B(A+B)^{-1}A+B(A+B)^{-1}B-A(A+B)^{-1}A$$ In a similar manner from $I=(A+B)^{-1}(A+B)$, we obtain $$\label{eq:matrix5}A-B=A(A+B)^{-1}B-B(A+B)^{-1}A+A(A+B)^{-1}A-B(A+B)^{-1}B$$ \eqref{eq:matrix4}+\eqref{eq:matrix5} results $$A(A+B)^{-1}B=B(A+B)^{-1}A$$

A mathematician who Twitter username is Manifoldless beat me to it by a few minutes :). But not just that. His solution is shorter (so better) than mine: \begin{align*}A(A+B)^{-1}B&=(A+B-B)(A+B)^{-1}(A+B-A)\\&=[I-B(A+B)^{-1}](A+B-A)\\&=A+B-A-B+B(A+B)^{-1}A\\&=B(A+B)^{-1}A\end{align*}

# Determinants as Area and Volume

The Area of a Parallelogram

Let $v=(v_1,v_2)$ and $w=(w_1,w_2)$ be two linearly independent vectors in $\mathbb{R}^2$. Then they span a parallelogram as shown in Figure 1.

Figure 1. Parallelogram spanned by two vector v and w.

The area $A$ of the parallelogram is
\begin{align*}
A&=||v||||w||\sin\theta\\
&=||v\times w||\\
&=\left|\begin{array}{cc}
v_1 & v_2\\
w_1 & w_2
\end{array}\right|.
\end{align*}
In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the area.

Exercise. Given two vectors $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ in $\mathbb{R}^3$, show that $||v||||w||\sin\theta=||v\times w||$ where $\theta$ is the angle between $v$ and $w$.

Hint. Note that $||v||^2||w||^2\sin^2\theta=||v||^2||w||^2-(v\cdot w)^2$.

The Volume of a Parallelepiped

Let $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$, $w=(w_1,w_2,w_3)$ be three linearly independent vectors in $\mathbb{R}^3$. Then they span a parallelepiped as shown in Figure 2.

Figure 2. Parallelepiped spanned by vectors u, v and w.

The volume $V$ of the parallelepiped is
\begin{align*}
V&=||u||\cos\theta ||v\times w||\\
&=u\cdot (v\times w)\\
&=\left|\begin{array}{ccc}
u_1 & u_2 & u_3\\
v_1 & v_2 & v_3\\
w_1 & w_2 & w_3
\end{array}\right|.
\end{align*}
In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the volume.

# Inverse of a Matrix

Let $A$ be an $n\times n$ matrix with $\det A\ne 0$. (A square matrix whose determinant is not equal to $0$ is called non-singular.) Let $X=(x_{ij})$ be an unknown $n\times n$ matrix such that
$AX=I$. Then
$$x_{1j}A^1+\cdots+x_{nj}A^n=E^j.$$
This is a system of linear equations and as we studied here, it can be solved by Cramer’s Rule as
\begin{align*}
x_{ij}&=\frac{\det(A^1,\cdots,E^j,\cdots A^n)}{\det A}\\
&=\frac{1}{\det A}\left|\begin{array}{ccccc}
a_{11} & \cdots & 0 & \cdots & a_{1n}\\
\vdots & & \vdots & & \vdots\\
a_{j1} & \cdots & 1 & \cdots & a_{jn}\\
\vdots & & \vdots & & \vdots\\
a_{n1} & \cdots & 0 & \cdots & a_{nn}
\end{array}\right|\\
&=\frac{1}{\det A}(-1)^{i+j}\det(A_{ji})
\end{align*}
for $i=1,\cdots,n$. If we show that $XA=I$ as well, then $X$ would be the inverse $A^{-1}$ and
$$A^{-1}=\frac{1}{\det A}{}^t((-1)^{i+j}\det(A_{ij})).$$
$\det({}^tA)=\det A\ne 0$, so we can find an $n\times n$ matrix $Y$ such that ${}^tAY=I$. Taking transposes, we obtain ${}^tYA=I$. Now,
$$I={}^tYA={}^tYIA={}^tY(AX)A={}^tYA(XA)=XA.$$

Example. Let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ with $\det A=ad-bc\ne 0$. Then $A^{-1}$ is given by
$$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b\\ -c & a \end{pmatrix}.$$

Example. Find the inverse of the matrix
$$A=\begin{pmatrix} 3 & 1 & -2\\ -1 & 1 & 2\\ 1 & -2 & 1 \end{pmatrix}.$$

Solution. $\det A=16$. We find
\begin{align*}
A_{11}=\begin{pmatrix}
1 & 2\\
-2 & 1
\end{pmatrix},\ A_{12}=\begin{pmatrix}
-1 & 2\\
1 & 1
\end{pmatrix},\ A_{13}=\begin{pmatrix}
-1 & 1\\
1 & -2
\end{pmatrix},\\
A_{21}=\begin{pmatrix}
1 & -2\\
-2 & 1
\end{pmatrix},\ A_{22}=\begin{pmatrix}
3 & -2\\
1 & 1
\end{pmatrix},\ A_{23}=\begin{pmatrix}
3 & 1\\
1 & -2
\end{pmatrix},\\
A_{31}=\begin{pmatrix}
1 & -2\\
1 & 2
\end{pmatrix},\ A_{32}=\begin{pmatrix}
3 & -2\\
-1 & 2
\end{pmatrix},\ A_{33}=\begin{pmatrix}
3 & 1\\
-1 & 1
\end{pmatrix}.
\end{align*}
Hence by the formula we obtain
\begin{align*}
A^{-1}&=\frac{1}{16}{}^t\begin{pmatrix}
\det(A_{11}) & -\det(A_{12}) & \det(A_{13})\\
-\det(A_{21}) & \det(A_{22}) & -\det(A_{23})\\
\det(A_{31}) &-\det(A_{32}) & \det(A_{33})
\end{pmatrix}\\
&=\frac{1}{16}{}^t\begin{pmatrix}
5 & 3 & 1\\
3 & 5 & 7\\
4 & -4 & 4
\end{pmatrix}\\
&=\frac{1}{16}\begin{pmatrix}
5 & 3 & 4\\
3 & 5 &-4\\
1 & 7 & 4
\end{pmatrix}.
\end{align*}

We introduce the following theorem without proof:

Theorem. For any two $n\times n$ matrices $A$, $B$,
$$\det(AB)=\det A\det B.$$

As a special case, we obtain:

Corollary. For an invertible matrix $A$,
$$\det(A^{-1})=\frac{1}{\det A}.$$

Proof. $AA^{-1}=I$, so by the theorem
$$1=\det(AA^{-1})=\det A\det (A^{-1}).$$
Thus proving the formula for the inverse.

# Cramer’s Rule

Consider a system of $n$ linear equations in $n$ unknowns
$$x_1A^1+\cdots+x_nA^n=B,$$
where $x_1,\cdots,x_n$ are variables and $A^1,\cdots,A^n,B$ are column vectors of dimension $n$. Suppose that $\det(A^1,\cdots,A^n)\ne 0$. Recall that this condition ensures that the linear system has a unique solution as seen here. Let us consider the determinant of the matrix obtained by replacing $j$-th column $A^j$ by $B$. Then using the properties of determinants studied here
\begin{align*}
\det (A^1,\cdots,B,\cdots,A^n)&=\det (A^1,\cdots,x_1A^1+\cdots+x_nA^n,\cdots,A^n)\\
&=\det (A^1,\cdots,x_1A^1,\cdots,x^nA^n)+\cdots+\det (A^1,\cdots,x_jA^j,\cdots,x_nA^n)\\
&+\cdots+\det (A^1,\cdots,x_nA^n,\cdots,x_nA^n)\\
&=x_1\det (A^1,\cdots,A^1,\cdots,A^n)+\cdots+x_j\det (A^1,\cdots,A^j,\cdots,A^n)\\
&+\cdots+x_n\det (A^1,\cdots,A^n,\cdots,A^n).
\end{align*}
Two column vectors are the same in every term except the $j$-th term, and so every term except the $j$-th term. Hence, we have
$$\det (A^1,\cdots,B,\cdots,A^n)=x_j\det (A^1,\cdots,A^j,\cdots,A^n),$$
i.e.
\begin{align*}
x_j&=\frac{\det (A^1,\cdots,B,\cdots,A^n)}{\det (A^1,\cdots,A^j,\cdots,A^n)}\\
&=\frac{\left|\begin{array}{ccccc}
a_{11} & \cdots & b_1 & \cdots & a_{1n}\\
a_{21} & \cdots & b_2 & \cdots & a_{2n}\\
\vdots & &\vdots& &\vdots\\
a_{n1} & \cdots & b_n & \cdots & a_{nn}
\end{array}\right|}{\left|\begin{array}{ccccc}
a_{11} & \cdots & a_{1j} & \cdots & a_{1n}\\
a_{21} & \cdots & a_{2j} & \cdots & a_{2n}\\
\vdots & &\vdots& &\vdots\\
a_{n1} & \cdots & a_{nj} & \cdots & a_{nn}
\end{array}\right|}
\end{align*}
for $j=1,\cdots,n$. This is called the Cramer’s Rule.

Example. Solve the system of Linear equations:
\begin{align*}
3x+2y+4z&=1,\\
2x-y+z&=0,\\
x+2y+3z&=1.
\end{align*}

Solution. By the Caramer’s Rule, we have
$$x=\frac{\left|\begin{array}{ccc} 1 & 2 & 4\\ 0 & -1 & 1\\ 1 & 2 & 3 \end{array}\right|}{\left|\begin{array}{ccc} 3 & 2 & 4\\ 2 & -1 & 1\\ 1 & 2 & 3 \end{array}\right|}=-\frac{1}{5},\ y=\frac{\left|\begin{array}{ccc} 3 & 1 & 4\\ 2 & 0 & 1\\ 1 & 1 & 3 \end{array}\right|}{\left|\begin{array}{ccc} 3 & 2 & 4\\ 2 & -1 & 1\\ 1 & 2 & 3 \end{array}\right|}=0,\ z=\frac{\left|\begin{array}{ccc} 3 & 2 & 0\\ 2 & -1 & 1\\ 1 & 2 & 0 \end{array}\right|}{\left|\begin{array}{ccc} 3 & 2 & 4\\ 2 & -1 & 1\\ 1 & 2 & 3 \end{array}\right|}=\frac{2}{5}.$$

# The Rank of a Matrix 2: The Rank of a Matrix and Subdeterminants

A test for the linear dependence of vectors may be given in terms of determinant.

Theorem. Let $A^1,\cdots,A^n$ be column vectors of dimension $n$. They are linearly dependent if and only if
$$\det(A^1,\cdots,A^n)=0.$$

Corollary. If a system of $n$ linear equations in $n$ unknowns has a matrix of coefficients whose determinants is not 0, then this system has a unique solution.

Proof. A system of $n$ linear equations in $n$ unknowns may be written as
$$x_1A^1+\cdots+x_nA^n=B,$$
where $A^1,\cdots,A^n$ are the column vectors of dimension $n$ of the matrix of coefficients and $B$ is a column vector of dimension $n$. Since $\det(A^1,\cdots,A^n)\ne 0$, $A^1,\cdots,A^n$ are linearly independent by the theorem. So there exists a unique solution $x_1,\cdots,x_n$ of the system.

Since determinants can be used to test linear dependence , they can be also used to determine the rank of a matrix in stead of using row operations as seen here.

Example. Let
$$A=\begin{pmatrix} 3 & 1 & 2 & 5\\ 1 & 2 & -1 & 2\\ 1 & 1 & 0 & 1 \end{pmatrix}.$$
Since $A$ is a $3\times 4$ matrix, its rank is at most 3. If we can find three linearly independent column vectors, the rank is 3. In fact,
$$\left|\begin{array}{ccc} 1 & 2 & 5\\ 2 & -1 & 2\\ 1 & 0 & 1 \end{array}\right|=4.$$
So, the rank is exactly 3.

Example. Let
$$B=\begin{pmatrix} 3 & 1 & 2 & 5\\ 1 & 2 & -1 & 2\\ 4 & 3 & 1 & 7 \end{pmatrix}.$$
Every $3\times 3$ subdeterminant has value 0, so the rank of $B$ is at most 2. The first two rows of $B$. The first two rows are linearly independent since teh determinant
$$\left|\begin{array}{cc} 3 & 1\\ 1 & 2 \end{array}\right|$$
is not 0. Hence, the rank is 2.