Lesson: A Review of Matrices

Question 2

1 point

Transcript

In our attempts to solve spanning and linear independence problems in Rn, we created a new object known as a matrix. We went on to study many properties of matrices, but for now, let’s review the properties of a matrix that paralleled our definitions of Rn. We should, of course, start with the definition of a matrix.

A matrix is a rectangular array of numbers. We say that A is an m-by-n matrix when A has m rows and n columns, such as seen here.

Two matrices A and B are equal if and only if they have the same size—that is, the same number of rows and the same number of columns—and their corresponding entries are equal—that is, if aij = bij for all i from 1 to m and j from 1 to n.

We sometimes denote the ijth entry of a matrix A by capital-(A)ij. This is taken to mean the same thing as lowercase-aij.

Next, we look at the definitions of addition and scalar multiplication in matrices. Let A and B be m-by-n matrices. We define addition of matrices by saying that (A + B)ij = (A)ij + (B)ij—that is, that the ijth entry of A + B is the sum of the ijth entry of A with the ijth entry of B. Let A be an m-by-n matrix, and t in R be a scalar. We define the scalar multiplication of matrices by saying that (tA)ij = t(A)ij, or that is, that the ijth entry of tA is t times the ijth entry of A.

With these definitions in hand, we noticed that we get the exact same theorem of useful properties as we did for Rn. Theorem 3.1.1: Let A, B, and C be m-by-n matrices, and let s and t be real scalars. Then we have the following properties.

  1. A + B is an m-by-n matrix, so we are closed under addition.
  2. A + B = B + A, so addition is commutative.
  3. Quantity (A + B) + C = A + quantity (B + C), so addition is associative.
  4. There exists a matrix, denoted by 0-sub-m,n such that A + 0m,n = A. That is, we have a 0 matrix.
  5. For each matrix A, there exists an m-by-n matrix inverse-A with the property that A + (inverse-A) = 0m,n. This is the additive inverse property.
  6. sA is an m-by-n matrix, so we are closed under scalar multiplication.
  7. s(tA) = (st)A, so scalar multiplication is associative.
  8. (s + t) quantity times A = sA + tA. This is a distributive law.
  9. s times the quantity (A + B) = sA + sB. This is another distributive law. And,
  10. 1A = A. This is the scalar multiplicative identity.

We were also able to define spanning sets and linear independence in matrices much as we had done in Rn. Let B be the set {A1 through Ak} of m-by-n matrices. Then the span of B is defined as the set of all linear combinations of the matrices in the set B.

For example, to determine if the matrix [-1, 2; 2, 1] is in the span of this set of matrices, we need to see if there are scalars t1, t2, t3, and t4 that satisfy this equation, so that we can write [-1, 2; 2, 1] as a linear combination of the matrices in our set. Performing the addition on the left side of this equality, we see that we are looking for the matrix [-1, 2; 2, 1] to equal the matrix [t1 – 2t3 + 2t4, t1 + 3t2 + 4t3 + 2t4; 2t1 + 4t2 – 4t3 – 4t4, 2t1 – 3t2 – 5t3 + 3t4]. By the definition of equality, this means that we are looking for solutions to the following system of linear equations. We solve the system by row reducing its augmented matrix, as seen here. Our final matrix is in reduced row echelon form. From this form, we see that we must have that t1 = 3, t2 = -1, t3 = 1, and t4 = -1 as a solution to our system. This means that we can write [-1, 2; 2, 1] as the linear combination 3[1, 1; 2, 2] – [0, 3; 4, -3] + [-2, 4; -4, -5] – [2, 2; -4, 3]. And thus, we’ve shown that [-1, 2; 2, 1] is in the span of our set.

After span, we looked at linear independence. Let B be the set {A1 through Ak} of m-by-n matrices. Then B is said to be linearly independent if the only solution to the equation t1A1 + through to tkAk = the 0 matrix is the trivial solution t1 through tk = 0. Otherwise, B is said to be linearly dependent.

For example, let’s determine whether or not this set is linearly independent. To do this, we need to see how many solutions there are to this equation, where we take a linear combination and set it equal to 0. Performing the calculations on the left-hand side, we see that this is the same as trying to see if the matrix whose entries are t1 + 8t3, 3t1 – 2t2 + 6t3, -t1 + t2 + 5t3, -3t1 + 5t2 – t3 is equal to the matrix [0, 0; 0, 0]. Now, this is the same as looking for solutions to the following system of homogeneous equations. We solve this system by row reducing its coefficient matrix, as seen here. The final matrix is in row echelon form, and so we see that the rank of the coefficient matrix is 3. Since this is the same as the number of variables, there are no parameters in the general solution to our homogeneous system. That means that there is only one solution to the system, and we know that this must be the trivial solution, and this means that our set is linearly independent.

Now let’s look at another example. Determine whether or not this set is linearly independent. Again, this is the same as looking for the number of solutions to the equation t1[1, 1; 3, -1] + t2[-1, 2; -8, 3] + t3[2, 11; -9, 4] = the 0 matrix. We perform the calculation on the left side; we see that this is the same as this matrix equality, and determining whether these matrices are equal is the same as looking for solutions to this system of homogeneous equations. We solve this system by row reducing the coefficient matrix, as seen here. Now, the final matrix is in row echelon form, so that we see that the rank of the coefficient matrix is 2, but since the number of variables in the system is 3, this means that there are 3 – 2 = 1 parameters in the general solution to the system. So, t1 = t2 = t3 = 0 is not the only solution to our system, and this means that our set is linearly dependent, or that is, it is not linearly independent.

© University of Waterloo and others, Powered by Maplesoft