Week 9

Determinant

Recall, for a $2\times2$ matrix $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant of $A$ is:

$$\det A = \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$$

And that $A^{-1}$ exists if and only if $\det A\neq 0$ (because the scalar coefficient is $\displaystyle\frac{1}{\det A}=\frac{1}{ad-bc}$).

And recall that to compute eigenvalues for a vector, we needed to find values of $\lambda$ such that $A\vector{v}=\lambda\vector{v}$ (where $\vector{v}\neq\vector{0}$).

So, the goal here is to find $\det A$ for some $n\times n$ matrix $A$.

Multilinearity

Take a unit cube, composed of three standard bases: $\vector{e}_1, \vector{e}_2, \vector{e}_3$.

We know that the volume of the unit cube is just $\text{length}\times\text{width}\times\text{height}$.

Or:

$$1 \times 1 \times 1$$

In other words, the volume $V$, is:

$$V = \det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix} = 1.$$

Then, if one of the side is of length $2$ e.g., take $2\vector{e}_2$. Then, the volume is now

$$2\times1\times1 = 2.$$

Or:

$$V = \det\begin{pmatrix} 2\vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix} = 2\det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix} = 2.$$

Then, for sides $a,b,c$, the volume of a given cube can be written as

$$V = \det\begin{pmatrix} a\vector{e}_1 \\ b\vector{e}_2 \\ c\vector{e}_3 \end{pmatrix}.$$

And by linearity (and that $\det I = 1$),

$$V = \det\begin{pmatrix} a\vector{e}_1 \\ b\vector{e}_2 \\ c\vector{e}_3 \end{pmatrix} = abc\cdot\det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix} = abc (1) = abc.$$

Also note that

$$\det\begin{pmatrix} a\vector{e}_1 \\ b\vector{e}_2 \\ c\vector{e}_3 \end{pmatrix} = \det\begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix} = abc$$

And so, we can surmise for any diagonal matrices, we have:

Antisymmetry

Again, consider a unit cube in $\R^3$. We now know that the determinant of the three basis vectors produces the volume, such as:

$$V = \det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix} = 1.$$

We also know that for a cube with sides of length one, its volume is always one. So, of course the order of the sides shouldn’t matter. However, by antisymmetry, switching two rows would mean:

$$V = -\det\begin{pmatrix} \vector{e}_2 \\ \vector{e}_1 \\ \vector{e}_3 \end{pmatrix} = -1.$$

The negative sign actually just indicates the orientation of the unit cube. Strictly speaking, the volume of the cube is actually the absolute value of the determinant:

$$V = \left|\det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix}\right|$$

And so, switching the rows in any order would not change the volume of the cube. In $\R^3$, there are $3! = 6$ combinations you can switch. Namely:

$$\det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_2 \\ \vector{e}_3 \end{pmatrix} = \det\begin{pmatrix} \vector{e}_3 \\ \vector{e}_1 \\ \vector{e}_2 \end{pmatrix} = \det\begin{pmatrix} \vector{e}_2 \\ \vector{e}_3 \\ \vector{e}_1 \end{pmatrix} = 1 \\ \det\begin{pmatrix} \vector{e}_2 \\ \vector{e}_1 \\ \vector{e}_3 \end{pmatrix} = \det\begin{pmatrix} \vector{e}_3 \\ \vector{e}_2 \\ \vector{e}_1 \end{pmatrix} = \det\begin{pmatrix} \vector{e}_1 \\ \vector{e}_3 \\ \vector{e}_2 \end{pmatrix} = -1$$

Now consider a case if we have two identical rows. Then, by this property, switching them would mean that:

$$\det\begin{pmatrix} \vector{v} \\ \vdots \\ \vector{v} \end{pmatrix} = -\det\begin{pmatrix} \vector{v} \\ \vdots \\ \vector{v} \end{pmatrix}$$

Remember that the determinant is just a number. So, the above is only true if and only if the determinant is zero.

$$\det\begin{pmatrix} \vector{v} \\ \vdots \\ \vector{v} \end{pmatrix} = -\det\begin{pmatrix} \vector{v} \\ \vdots \\ \vector{v} \end{pmatrix} \iff \det\begin{pmatrix} \vector{v} \\ \vdots \\ \vector{v} \end{pmatrix} = 0$$

Additionally, by properties 1 and 5:

$$\begin{align*} \det\begin{pmatrix} \vector{v}_1 \\ \vdots \\ \vector{v}_2 \end{pmatrix} &= \det\begin{pmatrix} \vector{v}_1 \\ \vdots \\ \vector{v}_2 \end{pmatrix} + \det\begin{pmatrix} c\vector{v}_2 \\ \vdots \\ \vector{v}_2 \end{pmatrix} \\ &= \det\begin{pmatrix} \vector{v}_1 \\ \vdots \\ \vector{v}_2 \end{pmatrix} + c\cdot\det\begin{pmatrix} \vector{v}_2 \\ \vdots \\ \vector{v}_2 \end{pmatrix} \\ &= \det\begin{pmatrix} \vector{v}_1 + c\vector{v}_2 \\ \vdots \\ \vector{v}_2 \end{pmatrix} \end{align*}$$

Which gives us

Diagonal matrices

The derivation for property 4 tells us that:

$$\det\begin{pmatrix} \lambda_1 & & 0 \\ & \ddots & \\ 0 & & \lambda_n \end{pmatrix} = \lambda_1\lambda_2\cdots\lambda_n.$$

However, this also holds for upper and lower triangular matrices.

$$\det\begin{pmatrix} \lambda_1 & & 0 \\ & \ddots & \\ * & & \lambda_n \end{pmatrix} = \det\begin{pmatrix} \lambda_1 & & * \\ & \ddots & \\ 0 & & \lambda_n \end{pmatrix} = \lambda_1\lambda_2\cdots\lambda_n.$$

Because trust me bro, that’s how it works. Or here, okay, I give a proof but I cheat you a little bit.

$$\begin{align*} \det\begin{pmatrix} \lambda_1 & a & b \\ 0 & \lambda_2 & c \\ 0 & 0 & \lambda_3 \end{pmatrix} &= \lambda_1 \lambda_2 \lambda_3 \det\begin{pmatrix} 1 & \frac{a}{\lambda_1} & \frac{b}{\lambda_1} \\ 0 & 1 & \frac{c}{\lambda_1} \\ 0 & 0 & 1 \end{pmatrix} \\ &= \lambda_1 \lambda_2 \lambda_3 \det\begin{pmatrix} 1 & \frac{a}{\lambda_1} & \frac{b}{\lambda_1} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} & \boxed{R_2 - \frac{c}{\lambda_1}R_3} \\ &= \lambda_1 \lambda_2 \lambda_3 \det\begin{pmatrix} 1 & \frac{a}{\lambda_1} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} & \boxed{R_1 - \frac{b}{\lambda_1}R_3} \\ &= \lambda_1 \lambda_2 \lambda_3 \det\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} & \boxed{R_1 - \frac{b}{\lambda_1}R_2} \\ &= \lambda_1 \lambda_2 \lambda_3 \end{align*}$$

By induction, this works for $n$ lambdas. And the same will apply for lower triangle matrices, just upside-down.

Applying different properties…

Consider a $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. We know that its determinant is $ad-bc$. We can also derive this by applying the properties of the determinant.

$$\begin{align*} \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \det\begin{pmatrix} a \vector{\mathbf{e}}_1 + b\vector{\mathbf{e}}_2 \\ c \vector{\mathbf{e}}_1 + d \vector{\mathbf{e}}_2 \end{pmatrix} \\ &= \det\begin{pmatrix} a \vector{\mathbf{e}}_1 \\ c \vector{\mathbf{e}}_1 \end{pmatrix} + \det\begin{pmatrix} a \vector{\mathbf{e}}_1 \\ d \vector{\mathbf{e}}_2 \end{pmatrix} + \det\begin{pmatrix} b \vector{\mathbf{e}}_2 \\ c \vector{\mathbf{e}}_1 \end{pmatrix} + \det\begin{pmatrix} b \vector{\mathbf{e}}_2 \\ d \vector{\mathbf{e}}_2 \end{pmatrix} \\ &= ac \det\begin{pmatrix} \vector{\mathbf{e}}_1 \\ \vector{\mathbf{e}}_1 \end{pmatrix} + ad \det\begin{pmatrix} \vector{\mathbf{e}}_1 \\ \vector{\mathbf{e}}_2 \end{pmatrix} + bc \det\begin{pmatrix} \vector{\mathbf{e}}_2 \\ \vector{\mathbf{e}}_1 \end{pmatrix} + bd \det\begin{pmatrix} \vector{\mathbf{e}}_2 \\ \vector{\mathbf{e}}_2 \end{pmatrix} \\ &= ac (0) + ad (1) + bc (-1) + bd (0) \\ &= ad - bc \end{align*}$$

We can also do the same thing for a $3\times3$ matrix $\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$.

$$\begin{align*} \det\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} &= \det\begin{pmatrix} a_{11}\vector{\mathbf{e}}_1 & a_{12}\vector{\mathbf{e}}_2 & a_{13}\vector{\mathbf{e}}_3 \\ a_{21}\vector{\mathbf{e}}_1 & a_{22}\vector{\mathbf{e}}_2 & a_{23}\vector{\mathbf{e}}_3 \\ a_{31}\vector{\mathbf{e}}_1 & a_{32}\vector{\mathbf{e}}_2 & a_{33}\vector{\mathbf{e}}_3 \end{pmatrix} \\ &= \text{decomposes to 27 terms...} \end{align*}$$

Notice how, for the $2\times2$, two of the terms cancel out to zero. Here, for a $3\times3$, only six terms survives. The determinant of the remaining 21 terms will be zero.

$$\begin{split} \det\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} &= a_{11} a_{12} a_{13} \det\begin{pmatrix} \vector{\mathbf{e}}_1 \\ \vector{\mathbf{e}}_2 \\ \vector{\mathbf{e}}_3 \end{pmatrix} + a_{11} a_{23} a_{32} \det\begin{pmatrix} \vector{\mathbf{e}}_1 \\ \vector{\mathbf{e}}_3 \\ \vector{\mathbf{e}}_2 \end{pmatrix} + a_{12} a_{21} a_{33} \det\begin{pmatrix} \vector{\mathbf{e}}_2 \\ \vector{\mathbf{e}}_1 \\ \vector{\mathbf{e}}_3 \end{pmatrix} \\ &\quad+ a_{12} a_{23} a_{31} \det\begin{pmatrix} \vector{\mathbf{e}}_2 \\ \vector{\mathbf{e}}_3 \\ \vector{\mathbf{e}}_1 \end{pmatrix} + a_{13} a_{21} a_{32} \det\begin{pmatrix} \vector{\mathbf{e}}_3 \\ \vector{\mathbf{e}}_1 \\ \vector{\mathbf{e}}_2 \end{pmatrix} + a_{13} a_{22} a_{31} \det\begin{pmatrix} \vector{\mathbf{e}}_3 \\ \vector{\mathbf{e}}_2 \\ \vector{\mathbf{e}}_1 \end{pmatrix} \end{split}$$

Then, after some careful calculation using the properties, we will see that the we end up with our expression for the determinant for a $3\times3$ matrix.

$$\begin{align*} \det\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} &= a_{11} \det\begin{pmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} - a_{12} \det\begin{pmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} + a_{13} \det\begin{pmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} \end{align*}$$

Then, what about $\R^4$ and beyond… the number of terms that it come out of matrix will grow exponentially. In fact, the only way to do this in a sane manner, is to turn to our good friend Gauss.