By Hefferon J.

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**Extra resources for A short linear algebra book (answers)**

**Example text**

The standard basis for R does not split into bases for the complementary subspaces the line x = y and the line x = −y. 43 (a) Yes, W1 +W2 = W2 +W1 for all subspaces W1 , W2 because each side is the span of W1 ∪W2 = W2 ∪W1 . (b) This one is similar to the prior one — each side of that equation is the span of (W1 ∪ W2 ) ∪ W3 = W1 ∪ (W2 ∪ W3 ). (c) Because this is an equality between sets, we can show that it holds by mutual inclusion. Clearly W ⊆ W + W . For W + W ⊆ W just recall that every subset is closed under addition so any sum of the form w1 + w2 is in W .

Thus the dimension is three. 17 (a) As in the prior exercise, the space M2×2 of matrices without restriction has this basis 1 0 0 0 , 0 0 1 0 0 0 0 , , 0 1 0 0 1 and so the dimension is four. (b) For this space { a b c d a = b − 2c and d ∈ R} = {b · 1 1 −2 0 0 0 +c· +d· 0 0 1 0 0 1 b, c, d ∈ R} this is a natural basis. 1 1 −2 0 0 , , 0 0 1 0 0 0 1 The dimension is three. (c) Gauss’ method applied to the two-equation linear system gives that c = 0 and that a = −b. Thus, we have this description { −b 0 b d b, d ∈ R} = {b · −1 1 0 +d· 0 0 0 and so this is a natural basis.

Remark. We can’t simply say that all of the rows are multiples of the first because the first row might be the zero row. Another Remark. 24 The column rank is two. One way to see this is by inspection — the column space consists of two-tall columns and so can have a dimension of at least two, and we can easily find two columns that together form a linearly independent set (the fourth and fifth columns, for instance). Another way to see this is to recall that the column rank equals the row rank, and to perform Gauss’ method, which leaves two nonzero rows.

### A short linear algebra book (answers) by Hefferon J.

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