## Linear Algebra |
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## Richard Kaye and Robert Wilson |

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.'
**

- 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