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Properties of Determinants of Matrices

We have seen how to compute the determinant of a matrix, and the incredible fact that we can perform expansion about any row or column to make this computation.

Properties of Determinants of Matrices

We have seen how to compute the determinant of a matrix, and the incredible fact that we can perform expansion about any row or column to make this computation. In this largely theoretical section, we will state and prove several more intriguing properties about determinants. Our main goal will be the two results in and , but more specifically, we will see how the value of a determinant will allow us to gain insight into the various properties of a square matrix.

A&= \begin{bmatrix} 0 & 8 & 3 & -4 \\ -1 & 2 & -2 & 5 \\ -2 & 8 & 4 & 3 \\ 0 & -4 & 2 & -3 \end{bmatrix} & B&= \begin{bmatrix} 0 & 8 & 3 & -4 \\ 0 & -4 & 2 & -3 \\ -2 & 8 & 4 & 3 \\ -1 & 2 & -2 & 5 \end{bmatrix}
State the theorem that allows us to make yet another extension to our NMEx series of theorems.
What is amazing about the interaction between matrix multiplication and the determinant?

Determinants and Row Operations

We start easy with a straightforward theorem whose proof presages the style of subsequent proofs in this subsection.

So tells us the effect of the first row operation () on the determinant of a matrix. Here is the effect of the second row operation.

Theorem: Determinant for Row or Column Multiples

Suppose that $A$ is a square matrix. Let $B$ be the square matrix obtained from $A$ by multiplying a single row by the scalar $\alpha$, or by multiplying a single column by the scalar $\alpha$. Then $\detname{B}=\alpha\detname{A}$.

Proof

Suppose that $A$ is a square matrix of size $n$ and we form the square matrix $B$ by multiplying each entry of row $i$ of $A$ by $\alpha$. Notice that the other rows of $A$ and $B$ are equal, so $\submatrix{A}{i}{j}=\submatrix{B}{i}{j}$, for all $1\leq j\leq n$. We compute $\detname{B}$ via expansion about row $i$. \detname{B} &= \sum_{j=1}^{n}(-1)^{i+j}\matrixentry{B}{ij}\detname{\submatrix{B}{i}{j}} &&\\ &= \sum_{j=1}^{n}(-1)^{i+j}\matrixentry{B}{ij}\detname{\submatrix{A}{i}{j}} &&\text{Hypothesis}\\ &= \sum_{j=1}^{n}(-1)^{i+j}\alpha\matrixentry{A}{ij}\detname{\submatrix{A}{i}{j}} &&\text{Hypothesis}\\ &= \alpha\sum_{j=1}^{n}(-1)^{i+j}\matrixentry{A}{ij}\detname{\submatrix{A}{i}{j}}\\ &= \alpha\detname{A}&&\\

The proof for the case of a multiple of a column is entirely similar, or could be derived from an application of employing the transpose of the matrix.

Let us go for understanding the effect of all three row operations. But first we need an intermediate result, but it is an easy one.

Now explain the third row operation. Here we go.

Is this what you expected? We could argue that the third row operation is the most popular, and yet it has no effect whatsoever on the determinant of a matrix! We can exploit this, along with our understanding of the other two row operations, to provide another approach to computing a determinant. We'll explain this in the context of an example.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

Determinants, Row Operations, Elementary Matrices

As a final preparation for our two most important theorems about determinants, we prove a handful of facts about the interplay of row operations and matrix multiplication with elementary matrices with regard to the determinant. But first, a simple, but crucial, fact about the identity matrix.

Theorem: Determinant of the Identity Matrix

For every $n\geq 1$, $\detname{I_n}=1$.

Proof

It may be overkill, but this is a good situation to run through a proof by induction on $n$ (). Is the result true when $n=1$? Yes, \detname{I_1} &=\matrixentry{I_1}{11}&&\\ &=1&&\\

Now assume the theorem is true for the identity matrix of size $n-1$ and investigate the determinant of the identity matrix of size $n$ with expansion about row 1, \detname{I_n} &= \sum_{j=1}^{n}(-1)^{1+j}\matrixentry{I_n}{1j}\detname{\submatrix{I_n}{1}{j}} &&\\ &= (-1)^{1+1}\matrixentry{I_n}{11}\detname{\submatrix{I_n}{1}{1}}\\ &\quad\quad+ \sum_{j=2}^{n}(-1)^{1+j}\matrixentry{I_n}{1j}\detname{\submatrix{I_n}{1}{j}}\\ &= 1\detname{I_{n-1}}+ \sum_{j=2}^{n}(-1)^{1+j}\,0\,\detname{\submatrix{I_n}{1}{j}} &&\\ &= 1(1)+\sum_{j=2}^{n}\,0=1 &&\text{Induction Hypothesis}\\

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

Determinants, Nonsingular Matrices, Matrix Multiplication

If you asked someone with substantial experience working with matrices about the value of the determinant, they'd be likely to quote the following theorem as the first thing to come to mind.

Theorem: Singular Matrices have Zero Determinants

Let $A$ be a square matrix. Then $A$ is singular if and only if $\detname{A}=0$.

Proof

Rather than jumping into the two halves of the equivalence, we first establish a few items. Let $B$ be the unique square matrix that is row-equivalent to $A$ and in reduced row-echelon form (, ). For each of the row operations that converts $B$ into $A$, there is an elementary matrix $E_i$ which effects the row operation by matrix multiplication (). Repeated applications of allow us to write A=E_sE_{s-1}\dots E_2E_1B

Then \detname{A} &= \detname{E_sE_{s-1}\dots E_2E_1B}\\ &=\detname{E_s}\detname{E_{s-1}}\dots\detname{E_2}\detname{E_1}\detname{B} &&

From we can infer that the determinant of an elementary matrix is never zero (note the ban on $\alpha=0$ for $\elemmult{\alpha}{i}$ in ). So the product on the right is composed of nonzero scalars, with the possible exception of $\detname{B}$. More precisely, we can argue that $\detname{A}=0$ if and only if $\detname{B}=0$. With this established, we can take up the two halves of the equivalence.

If $A$ is singular, then by , $B$ cannot be the identity matrix. Because (1) the number of pivot columns is equal to the number of nonzero rows, (2) not every column is a pivot column, and (3) $B$ is square, we see that $B$ must have a zero row. By the determinant of $B$ is zero, and by the above, we conclude that the determinant of $A$ is zero.

We will prove the contrapositive (). So assume $A$ is nonsingular, then by , $B$ is the identity matrix and tells us that $\detname{B}=1\neq 0$. With the argument above, we conclude that the determinant of $A$ is nonzero as well.

For the case of $2\times 2$ matrices you might compare the application of with the combination of the results stated in and .

Example: Zero and nonzero determinant, Archetypes A and B

The coefficient matrix in has a zero determinant (check this!) while the coefficient matrix has a nonzero determinant (check this, too). These matrices are singular and nonsingular, respectively. This is exactly what says, and continues our list of contrasts between these two archetypes.

In we said singular matrices are a distinct minority. If you built a random matrix and took its determinant, how likely would it be that you got zero?

Since is an equivalence () we can expand on our growing list of equivalences about nonsingular matrices. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants.

Condensed: the full section is in Beezer, A First Course in Linear Algebra.

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Pertanyaan yang sering diajukan

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn: it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

Bagian dari halaman ini diadaptasi dari Beezer, A First Course in Linear Algebra (GFDL 1.2). Diringkas dan dijelaskan ulang di sini; kesalahan adalah tanggung jawab kami.

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