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Chapter 2: Problem 10

Let \(A\) be an \(n \times n\) matrix. Prove that \(A\) is a diagonal matrix if and only if \(A_{i j}=\delta_{i j} A_{i j}\) for all \(i\) and \(j\).

Short Answer

Expert verified
In summary, we proved that an $n \times n$ matrix $A$ is a diagonal matrix if and only if \(A_{i j}=\delta_{i j} A_{i j}\) for all \(i\) and \(j\), by showing that if \(A\) is a diagonal matrix, then \(A_{i j} = \delta_{i j} A_{i j}\) for all \(i\), \(j\) (Part 1), and if \(A_{i j} = \delta_{i j} A_{i j}\) for all \(i\), \(j\), then \(A\) is a diagonal matrix (Part 2). This result relies on understanding of diagonal matrices and the Kronecker delta function.

Step by step solution

01

Part 1: If A is a diagonal matrix, then \(A_{i j} = \delta_{i j} A_{i j}\) for all i, j

Assume that \(A\) is a diagonal matrix. Then, its elements \(A_{ij}\) must follow the main diagonal constraint: \[A_{ij} = \left\{ \begin{array}{ll} a_{ii} & \mbox{if } i = j \\ 0 & \mbox{if } i \neq j \end{array} \right. \] Now, we determine \(A_{ij}\) in terms of the Kronecker delta function: \[A_{ij} = \delta_{ij}A_{ij} = \left\{ \begin{array}{ll} 1 \cdot a_{ii} & \mbox{if } i = j \\ 0 \cdot a_{ij} & \mbox{if } i \neq j \end{array} \right. \] This is true for all \(i\) and \(j\). Therefore, if \(A\) is a diagonal matrix, then \(A_{ij} = \delta_{ij}A_{ij}\) for all \(i\), \(j\).
02

Part 2: If A_ij=未_ij A_ij for all i, j, then A is a diagonal matrix

Assume that \(A_{ij}=\delta_{ij}A_{ij}\) for all \(i\), \(j\). We rewrite this equation as: \[A_{ij} = \left\{ \begin{array}{ll} a_{ii} & \mbox{if } i = j \\ 0 & \mbox{if } i \neq j \end{array} \right. \] Since \(A_{ij}\) is zero for all \(i \neq j\), and only the elements \(a_{ii}\) are nonzero, this means that \(A\) is a diagonal matrix. Hence, if \(A_{ij} = \delta_{ij}A_{ij}\) for all \(i\), \(j\), then \(A\) is a diagonal matrix.
03

Conclusion

By proving Part 1 and Part 2, we have shown that \(A\) is a diagonal matrix if and only if \(A_{ij} = \delta_{ij}A_{ij}\) for all \(i\) and \(j\).

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Most popular questions from this chapter

Suppose \(A\) and \(B\) are orthogonal matrices. Show that \(A^{T}, A^{-1}, A B\) are also orthogonal.

Let \(\beta\) and \(\gamma\) be the standard ordered bases for \(\mathrm{R}^{n}\) and \(\mathrm{R}^{m}\), respectively. For each linear transformation \(\mathrm{T}: \mathrm{R}^{n} \rightarrow \mathrm{R}^{m}\), compute \([\mathrm{T}]_{\beta}^{\gamma}\). (a) \(\mathrm{T}: \mathrm{R}^{2} \rightarrow \mathrm{R}^{3}\) defined by $\mathrm{T}\left(a_{1}, a_{2}\right)=\left(2 a_{1}-a_{2}, 3 a_{1}+4 a_{2}, a_{1}\right)$. (b) \(\mathrm{T}: \mathrm{R}^{3} \rightarrow \mathrm{R}^{2}\) defined by $\mathrm{T}\left(a_{1}, a_{2}, a_{3}\right)=\left(2 a_{1}+3 a_{2}-a_{3}, a_{1}+a_{3}\right)$. (c) \(\mathrm{T}: \mathrm{R}^{3} \rightarrow R\) defined by \(\mathrm{T}\left(a_{1}, a_{2}, a_{3}\right)=2 a_{1}+a_{2}-3 a_{3}\). (d) \(\mathrm{T}: \mathrm{R}^{3} \rightarrow \mathrm{R}^{3}\) defined by $$ \mathrm{T}\left(a_{1}, a_{2}, a_{3}\right)=\left(2 a_{2}+a_{3},-a_{1}+4 a_{2}+5 a_{3}, a_{1}+a_{3}\right) . $$ (e) \(\mathrm{T}: \mathrm{R}^{n} \rightarrow \mathrm{R}^{n}\) defined by $\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=\left(a_{1}, a_{1}, \ldots, a_{1}\right)$. (f) \(\mathrm{T}: \mathrm{R}^{n} \rightarrow \mathrm{R}^{n}\) defined by $\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=\left(a_{n}, a_{n-1}, \ldots, a_{1}\right)$. (g) \(\mathrm{T}: \mathrm{R}^{n} \rightarrow R\) defined by \(\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1}+a_{n}\).

Let \(A=\left[\begin{array}{ll}1 & 2 \\ 3 & 6\end{array}\right] .\) Find a \(2 \times 3\) matrix \(B\) with distinct nonzero entries such that \(A B=0.\)

Which of the following matrices are normal? \(A=\left[\begin{array}{rr}3 & -4 \\\ 4 & 3\end{array}\right], B=\left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right], C=\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\)

Partition each of the following matrices so that it becomes a square block matrix with as many diagonal blocks as possible: \\[ A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 3 \end{array}\right], \quad B=\left[\begin{array}{lllll} 1 & 2 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 6 \end{array}\right], \quad C=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{array}\right] \\]
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