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Linear Algebra, 2020a

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234 Chapter Three. Maps Between Spaces<br />

1.4 Theorem Let h, g: V → W be linear maps represented with respect to bases<br />

B, D by the matrices H and G and let r be a scalar. Then with respect to<br />

B, D the map r · h: V → W is represented by rH and the map h + g: V → W is<br />

represented by H + G.<br />

Proof Generalize the examples. This is Exercise 10.<br />

QED<br />

1.5 Remark These two operations on matrices are simple, but we did not define<br />

them in this way because they are simple. We defined them this way because<br />

they represent function addition and function scalar multiplication. That is,<br />

our program is to define matrix operations by referencing function operations.<br />

Simplicity is a bonus.<br />

We will see this again in the next subsection, where we will define the<br />

operation of multiplying matrices. Since we’ve just defined matrix scalar multiplication<br />

and matrix sum to be entry-by-entry operations, a naive thought is<br />

to define matrix multiplication to be the entry-by-entry product. In theory we<br />

could do whatever we please but we will instead be practical and combine the<br />

entries in the way that represents the function operation of composition.<br />

A special case of scalar multiplication is multiplication by zero. For any map<br />

0 · h is the zero homomorphism and for any matrix 0 · H is the matrix with all<br />

entries zero.<br />

1.6 Definition A zero matrix has all entries 0. We write Z n×m or simply Z<br />

(another common notation is 0 n×m or just 0).<br />

1.7 Example The zero map from any three-dimensional space to any twodimensional<br />

space is represented by the 2×3 zero matrix<br />

( )<br />

0 0 0<br />

Z =<br />

0 0 0<br />

no matter what domain and codomain bases we use.<br />

Exercises<br />

̌ 1.8 Perform<br />

(<br />

the indicated<br />

) (<br />

operations,<br />

)<br />

if defined, or state “not defined.”<br />

5 −1 2 2 1 4<br />

(a)<br />

+<br />

6 1 1 3 0 5<br />

( )<br />

2 −1 −1<br />

(b) 6 ·<br />

1 2 3<br />

( ) ( )<br />

2 1 2 1<br />

(c) +<br />

0 3 0 3<br />

( ) ( )<br />

1 2 −1 4<br />

(d) 4 + 5<br />

3 −1 −2 1

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