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Chapter 2: Problem 50

Determine whether the matrix is idempotent. A square matrix \(A\) is idempotent when \(A^{2}=A\). $$ \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] $$

Short Answer

Expert verified
Yes, the given matrix is idempotent.

Step by step solution

01

Identify the Matrix

The given matrix is \[ \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \]
02

Determine the Square of the Matrix

Matrix multiplication is performed by taking the dot products of the corresponding row and column vectors. In this case, the matrix's square \(A^{2}\) is computed by multiplying the matrix with itself. Thus, \[A^{2} = \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \times \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\] which results in \[\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\]
03

Compare \(A^{2}\) with Original Matrix \(A\)

The final step is to compare the result \(A^{2}\) with the original matrix \(A\). Since both are identical, we can now confirm that the given matrix is indeed idempotent.

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Most popular questions from this chapter

Proof Prove that if \(A^{2}=A\), then $$I-2 A=(I-2 A)^{-1}$$

Show that \(A C=B C\), even though \(A \neq B\). $$A=\left[\begin{array}{rrr}1 & 2 & 3 \\\0 & 5 & 4 \\\3 & -2 & 1\end{array}\right], \quad B=\left[\begin{array}{rrr}4 & -6 & 3 \\\5 & 4& 4 \\\\-1 & 0 & 1\end{array}\right]$$ $$C=\left[\begin{array}{rrr}0 & 0 & 0 \\\0 & 0 & 0 \\\4 & -2 & 3\end{array}\right]$$

Find the least squares regression line. $$ (0,0),(1,1),(2,4) $$

Perform the operations, given \(c=-2\) and \(A=\left[\begin{array}{rrr}1 & 2 & 3 \\\ 0 & 1 & -1\end{array}\right], B=\left[\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right], C=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\). $$c B(C+C)$$.

Prove that if \(A\) is an \(n \times n\) matrix, then \(A-A^{T}\) is skew-symmetric.
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