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Chapter 6: Problem 41

Find the nullity of \(T\) $$T: R^{4} \rightarrow R^{4}, \operatorname{rank}(T)=0$$

Short Answer

Expert verified
The nullity of the transformation T is 4.

Step by step solution

01

Understand the rank-nullity theorem and its application

The rank-nullity theorem states that Rank(T) + Nullity(T) = dim(T). It is also known as the dimension theorem for its relation between the rank, nullity, and the dimension of a linear transformation.
02

Use the given data and apply the theorem

The rank of the transformation T is given as 0, and the transformation is between spaces of dimension 4, ie. \(R^{4} \rightarrow R^{4}\). Hence, dim(T) = 4. Substitute these values into the rank-nullity theorem: 0 + Nullity(T) = 4.
03

Find the nullity

Solving the equation from Step 2 for Nullity(T) gives Nullity(T) = 4. Hence, the nullity of the transformation T is 4.

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Most popular questions from this chapter

Let \(A\) be an \(n \times n\) matrix such that \(A^{2}=O .\) Prove that if \(B\) is similar to \(A,\) then \(B^{2}=O\).

Give a geometric description of the linear transformation defined by the elementary matrix. $$A=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]$$

Let \(B=\\{(1,2),(-1,-1)\\}\) and \(B^{\prime}=\\{(-4,1),(0,2)\\}\) be bases for \(R^{2},\) and let \(A=\left[\begin{array}{rr}2 & 1 \\ 0 & -1\end{array}\right]\) be the matrix for \(T: R^{2} \rightarrow R^{2}\) relative to \(B\). (a) Find the transition matrix \(P\) from \(B^{\prime}\) to \(B\) (b) Use the matrices \(A\) and \(P\) to find \([\mathbf{v}]_{B}\) and \([T(\mathbf{v})]_{B},\) where \([\mathbf{v}]_{B^{\prime}}=\left[\begin{array}{r}-1 \\ 4\end{array}\right]\). (c) Find \(A^{\prime}\) (the matrix for \(T\) relative to \(B^{\prime}\) ) and \(P^{-1}\). (d) Find \([T(\mathbf{v})]_{B^{\prime}}\) in two ways: first as \(P^{-1}[T(\mathbf{v})]_{B}\) and then \(\operatorname{as} A^{\prime}[(\mathbf{v})]_{B^{\prime}}\)

Find \(T(\mathbf{v})\) by using (a) the standard matrix and (b) the matrix relative to \(B\) and \(B^{\prime}\). $$\begin{array}{l} T: R^{3} \rightarrow R^{4}, T(x, y, z)=(2 x, x+y, y+z, x+z) \\ \mathbf{v}=(1,-5,2) \\ B=\\{(2,0,1),(0,2,1),(1,2,1)\\} \\ B^{\prime}=\\{(1,0,0,1),(0,1,0,1),(1,0,1,0),(1,1,0,0)\\} \end{array}$$

Let \(B=P^{-1} A P .\) Prove that if \(A \mathbf{x}=\mathbf{x},\) then \(P B P^{-1} \mathbf{x}=\mathbf{x}\).
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