# Linear Algebra

## Richard Kaye and Robert Wilson

Reviews | Ordering information | Errata

## Information

Linear algebra, by RW Kaye and RA Wilson, can be ordered from the Internet bookshop, or from the publishers, Oxford University Press at £28.00 paperback (ISBN 0198502370).

## Reviews

From the review in the Mathematical Gazette, March 1999:
Familiarity is apt to dull the senses when reviewing mainstream texts such as this one, but make no mistake, this is a clearly and carefully written book that lecturers could easily lecture from and learners readily learn from: it merits a wide circulation.'

From the review in the Nieuw Archief voor Wiskunde, July 1999:
... this book can be strongly recommended.'

## Corrections

The following errors and misprints have already been found. If you have found any others, please email me: R.A.Wilson [At] qmul [Dot] ac [Dot] uk
• p. 10: the top right entries in the inverse of $$\mathbf A$$ on lines 4 and 6 should be $$-5/4$$ not $$-5/2$$.
• pp. 10-11: in the example on solving linear equations, the third equation should have $$-3$$, not 3. Similarly for the column vector at the very end of p. 10, and the last column of the first matrix on p. 11.
• p. 12: in the equation for the expansion of a determinant by its top row the entries $$a_{1n}$$ in the second row, and in the top row of the first two subdeterminants, should be $$a_{2n}$$, and in the last subdeterminant, $$a_{1\;n-1}$$ should be $$a_{2\; n-1}$$.
• p. 31: in line -3, replace $$+\cdots+$$ by $$-\cdots-$$.
• p. 67: the base change in Example 4.13 has been done in the wrong order. Replace the last 6 lines of the page by the following:

If $$(v_1,v_2)^T$$ is the coordinate form of a vector $$\mathbf v$$ with respect to $$\mathbf b_1,\mathbf b_2$$, then $$\mathbf Q(v_1,v_2)^T$$ is the coordinate form of $$\mathbf v$$ with respect to the usual basis. Then $$\mathbf P^{-1}\mathbf Q(v_1,v_2)^T$$ is the coordinate form of $$\mathbf v$$ with respect to $$\mathbf a_1,\mathbf a_2$$, so the base-change matrix from $$\mathbf a_1,\mathbf a_2$$ to $$\mathbf b_1,\mathbf b_2$$ is \begin{align*} \mathbf P^{-1}\mathbf Q &= \begin{pmatrix} 1 & 1 \\ i & 1+i \end{pmatrix}^{-1} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}= \begin{pmatrix} 1+i & -1 \\ -i & 1 \end{pmatrix} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}\\ &= \begin{pmatrix} -1+2i & i \\ 1-i & 1-i \end{pmatrix}. \end{align*}

• p. 72: in Exercise 4.6, the last occurrence of $$\mathbf f_1$$ should read $$\mathbf f_3$$ .
• p. 88: in the definition of $$v_S$$ replace the coefficients of the $$u_j$$ by their complex conjugates (equivalently, swap $$v$$ and $$u_j$$ in this inner product).
• p. 91: in line 12, replace $$P_{n-1}$$ by $$P_{n-1}(x)$$. Similarly in line 14, replace $$P_n$$ by $$P_n(x)$$.
• p. 115: in the line beginning $$\mathbf P^T=\ldots$$, the last entry of the first matrix should be $$1/\surd{11}$$, not $$\surd{11}$$.
• p. 117: in line 15, the second partial derivative should be $$\partial F/\partial y$$ not $$\partial F/\partial x$$.
• p. 121: in Proposition 7.24, add 'Hermitian matrix over' after 'or' in first line. Also add ', minus the number of $$i < n$$ such that $$D_i, D_{i+1}$$ have different sign' at the end of the statement.
• p. 140: in Exercise 8.4, replace $$-y+3z$$ by $$-y-2z$$ in the definition of $$f$$.
• p. 157: the second line of $$\mathbf P^{-1}$$ should be negated.
• p. 160: in line 14, it is not true that $$(\mathbf A-\lambda_n\mathbf I)\mathbf e_n = \mathbf 0$$. In fact it is a linear combination of $$\mathbf e_1, \ldots, \mathbf e_{n-1}$$. Thus the induction still works, but the proof needs some slight adjustments. For example, replace the third paragraph of the proof (beginning 'Now ...') by the following:

Now $$(\mathbf A-\lambda_n\mathbf I)\mathbf e_n$$ is a linear combination of $$\mathbf e_1, \ldots, \mathbf e_{n-1}$$. Also, the linear transformation given by $$\mathbf A$$ on the subspace $$\mathrm{span}(\mathbf e_1,\ldots,\mathbf e_{n-1})$$ has upper triangular matrix with respect to this basis, with diagonal entries $$\lambda_1,\ldots,\lambda_{n-1}$$, so by the induction hypothesis we have $(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I) \mathbf e_i =\mathbf 0$ for all $$i < n$$. Thus $(\mathbf A-\lambda_n\mathbf I) (\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I) \mathbf e_i =\mathbf 0$ for $$i < n$$. Moreover, $(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I) (\mathbf A-\lambda_n\mathbf I)\mathbf e_n$ is a linear combination of the$(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I) \mathbf e_i$ for $$i < n$$, all of which are $$\mathbf 0$$. Therefore $(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I) (\mathbf A-\lambda_n\mathbf I) \mathbf e_i =\mathbf 0$ for all $$i\le n$$.

• p. 184: $$\mathbf P = \begin{pmatrix}3& 1\cr 1& -1\end{pmatrix}$$, not $$\begin{pmatrix}3& -1\cr 1& -1\end{pmatrix}$$.
• p. 186: in Exercise 12.8(a), $$4x_n$$ should read $$4y_n$$.
• p. 209: in lines 3-4, read $$v_2=v_3=\cdots=v_k=0$$, so that $$\mathbf v = (1,0,...,0)^T$$.
• p. 226: remove the minus sign from $$-ih$$ in line 10 (twice) and line 14.

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Last content update 19 June 2007.
R.A.Wilson