is singular is nonsingular
Properties of the Determinant
We begin by summarizing the properties of determinants we introduced in previous sections.
- The determinant of the identity matrix is equal to 1. (Lemma lemma:detofid)
- The determinant of a triangular matrix is equal to the product of the main diagonal entries. (Lemma lemma:triangulardet)
- The determinant of the transpose is equal to the determinant of the matrix. (Theorem th:detoftrans)
- If a matrix contains a row of zeros, then its determinant is equal to 0. (Lemma lemma:det0lemma)
- If two rows of a matrix are the same, then the determinant of the matrix is equal to 0. (Lemma lemma:det0lemma)
- If one row of a matrix is a scalar multiple of another row, then the determinant of the matrix is equal to 0. (Lemma lemma:det0lemma)
- If is obtained from by interchanging two different rows, then
- If is obtained from by multiplying one of the rows of by a non-zero constant . Then
- If is obtained from by adding a multiple of one row of to another row, then
(The last three properties comprise Theorem th:elemrowopsanddet)
In this section we will prove the following important results:
- (a)
- A square matrix is singular if and only if its determinant is equal to 0.
- (b)
- The determinant of a product is the product of the determinants.
To get us started, we need the following lemma.
- Proof
- Recall that if is obtained from using an elementary row operation,
then the same elementary row operation carries to . There are three types of
elementary row operations and three types of elementary matrices, so we will
have to consider three cases.
Case 1. Suppose is obtained from by interchanging two rows, then so
Case 2. Suppose is obtained from by multiplying one of the rows of by a non-zero constant , then so
Case 3. Suppose is obtained from by adding a scalar multiple of one row to another row, then so
Invertibility and the Determinant
Recall that we first introduced determinants in the context of invertibility of matrices. Specifically, we found that is invertible if and only if . (A logically equivalent statement is: is singular if and only if .) We are now in the position to prove this result for all square matrices.
- Proof
- Let be a square matrix. To determine whether is singular we need to
find . In Elementary Matrices we found that there exist elementary matrices
such that
so
By repeated application of Lemma lemma:detelemproduct, we find that
Suppose that is singular, then . But then contains a row of zeros, and . (Lemma
lemma:det0lemma) Since determinants of elementary matrices are non-zero, we conclude that .
Conversely, suppose , then But then , so is singular.
Determine whether is an invertible matrix without using elementary row
operations.
Compute the determinant of . You will find that . By Theorem th:detofsingularmatrix we conclude that is
not invertible.
Determinant of a Product
- Proof
- Suppose is invertible, then can be written as a product of elementary
matrices. (Theorem th:elemmatrices)
Then, by repeated application of Lemma lemma:detelemproduct, we get \begin{align*} \det{AB}&=\det{(E_1E_2\ldots E_kB)}\\ &=\det{E_1}\det{E_2}\ldots \det{E_k}\det{B}\\ &=\det{(E_1E_2\ldots E_k)}\det{B}\\ &=\det{A}\det{B} \end{align*}
Now suppose that is not invertible. Then is also not invertible. So, and . Thus .
The following theorem is a nice consequence of Theorem th:detofproduct. We leave the proof to the reader. (Practice Problem prob:proofdetofinverse)
Practice Problems
Problems prob:singmatrixdet1a-prob:singmatrixdet1b
Without doing written computations, determine whether the given matrix is singular.
Find values of for which the given matrix is singular.
List values of in an increasing order.
Answer:
Problems prob:detproduct1a-prob:detproduct1c
Suppose and are matrices such that and . Find each of the following.
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