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Chapter 5: Problem 35

Let \(k\) be a nonzero scalar. Show that a linear map \(T\) is singular if and only if \(k T\) is singular. Hence, \(T\) is singular if and only if \(-T\) is singular.

Short Answer

Expert verified
A linear map \(T\) is singular if and only if \(kT\) is singular, where \(k\) is a nonzero scalar. We proved this by showing that if \(T\) is singular, then \(kT\) is singular, and if \(kT\) is singular, then \(T\) is singular. Moreover, by setting \(k=-1\), we can conclude that \(T\) is singular if and only if \(-T\) is singular.

Step by step solution

01

1. Proving \(T\) singular implies \(kT\) is singular

Assume that \(T\) is singular. To prove that \(kT\) is singular, we must show that there is no linear map \((kT)^{-1}\) such that \((kT)^{-1}(kT) = I_V\) and \((kT)(kT)^{-1} = I_W\). Suppose \((kT)^{-1}\) exists. Then, we have \[(kT)^{-1}(kT) = (kT)^{-1}(kI_VT) = (kI_V)^{-1}T^{-1}(kT).\] Since \(k\) is a nonzero scalar, \((kI_V)^{-1}\) exists, and using properties of the inverse we get: \[(kI_V)^{-1}T^{-1}(kT) = T^{-1}(kI_V)^{-1}(kT) = T^{-1}T = I_V,\] which implies that \(T^{-1}\) exists, contradicting the fact that \(T\) is singular. Therefore, \(kT\) must also be singular.
02

2. Proving \(kT\) singular implies \(T\) is singular

Assume that \(kT\) is singular. To prove that \(T\) is singular, we must show that there is no linear map \(T^{-1}\) such that \(T^{-1}T = I_V\) and \(TT^{-1} = I_W\). Suppose \(T^{-1}\) exists. Then, we can define a map \((kT)^{-1}\) as follows: \[(kT)^{-1} = \frac{1}{k} T^{-1}.\] Now, let's verify that \((kT)(kT)^{-1} = I_W\). \[(kT)(kT)^{-1} = (kT)(\frac{1}{k}T^{-1}) = kI_W T \frac{1}{k}T^{-1} = I_W T T^{-1} = I_W,\] and that \((kT)^{-1}(kT) = I_V\). \[(kT)^{-1}(kT) = \frac{1}{k}T^{-1}(kT) = T^{-1}T = I_V.\] Since \((kT)^{-1}\) exists, we have contradicted the fact that \(kT\) is singular. Therefore, \(T\) must also be singular.
03

Conclusion

We have shown that a linear map \(T\) is singular if and only if \(kT\) is singular, where \(k\) is a nonzero scalar. To show that \(T\) is singular if and only if \(-T\) is singular, we can set \(k=-1\), and the above results will hold for this case as well.

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Most popular questions from this chapter

Suppose that \(F: V \rightarrow U\) is linear and that \(V\) is of finite dimension. Show that \(V\) and the image of \(F\) have the same dimension if and only if \(F\) is nonsingular. Determine all nonsingular linear mappings \(T: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3}.\)

Let \(A\) be an \(n \times n\) matrix with characteristic polynomial $$ f(t)=(-1)^{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{1} t+a_{0} . $$ (a) Prove that \(A\) is invertible if and only if \(a_{0} \neq 0\). (b) Prove that if \(A\) is invertible, then $$ A^{-1}=\left(-1 / a_{0}\right)\left[(-1)^{n} A^{n-1}+a_{n-1} A^{n-2}+\cdots+a_{1} I_{n}\right] . $$ (c) Use (b) to compute \(A^{-1}\) for $$ A=\left(\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & -1 \end{array}\right). $$

Let \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) be the linear mapping defined by \\[ G(x, y, z)=(x+2 y-z, \quad y+z, \quad x+y-2 z) \\] Find a basis and the dimension of (a) the image of \(G,(\mathrm{b})\) the kernel of \(G\)

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Let \(A\) be an \(n \times n\) matrix that is similar to an upper triangular matrix and has the distinct eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\( with corresponding multiplicities \)m_{1}, m_{2}, \ldots, m_{k}$. Prove the following statements. (a) \(\operatorname{tr}(A)=\sum^{k} m_{i} \lambda_{i}\) (b) $\operatorname{det}(A)=\left(\lambda_{1}\right)^{m_{1}}\left(\lambda_{2}\right)^{m_{2}} \ldots\left(\lambda_{k}\right)^{m_{k}}$.
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