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91Ó°ÊÓ

True or false? (Explain your answer.) There exists \(r>0\) and a differentiable inap $$ g: B_{r}\left(\left[\begin{array}{rr} -3 & 0 \\ 0 & -3 \end{array}\right]\right) \rightarrow \operatorname{Mat}(2,2) \text { such that } g\left(\left[\begin{array}{rr} -3 & 0 \\ 0 & -3 \end{array}\right]\right)=\left[\begin{array}{rr} 1 & 2 \\ -2 & -1 \end{array}\right] $$ and \((g(A))^{2}=A\) for all \(A \in B_{r}\left(\left[\begin{array}{rr}-3 & 0 \\\ 0 & -3\end{array}\right]\right)\)

Short Answer

Expert verified
True, the function \( g(A) \) satisfies \((g(A))^2 = A\).

Step by step solution

01

Identify Given Matrix and Neighborhood

The given matrix \( A \) is \( A = \begin{bmatrix}-3 & 0 \ 0 & -3\end{bmatrix} \). We have an open neighborhood around \( A \) defined by the radius \( r > 0 \), noted as \( B_r(A) \). The function \( g \) is supposed to be differentiable on this neighborhood, \( B_r(A) \), and maps to \( \operatorname{Mat}(2,2) \). We know that \( g(A) = \begin{bmatrix}1 & 2 \ -2 & -1\end{bmatrix} \).
02

Evaluate Condition \((g(A))^2 = A\)

The condition \((g(A))^2 = A\) must hold for all \( A \) in \( B_r(A) \). First, calculate \((g(A))^2\). Given \( g(A) = \begin{bmatrix}1 & 2 \ -2 & -1\end{bmatrix} \), then:\[(g(A))^2 = \begin{bmatrix}1 & 2 \ -2 & -1\end{bmatrix} \begin{bmatrix}1 & 2 \ -2 & -1\end{bmatrix} = \begin{bmatrix}1\times1 + 2\times(-2) & 1\times2 + 2\times(-1) \-2\times1 + (-1)\times(-2) & -2\times2 + (-1)\times(-1)\end{bmatrix} = \begin{bmatrix}-3 & 0 \ 0 & -3\end{bmatrix}.\]This is equal to \( A \).
03

Interpretation and Conclusion

Since for the specified matrix \( A = \begin{bmatrix}-3 & 0 \ 0 & -3\end{bmatrix} \), \((g(A))^2 = A\) is satisfied with \( g \left( \begin{bmatrix}-3 & 0 \ 0 & -3\end{bmatrix} \right) = \begin{bmatrix}1 & 2 \ -2 & -1\end{bmatrix} \), we evaluate the differentiability of \( g \) in the neighborhood. Because \( g \) produces a correct square root for \( A \), it implies this can be the case within the neighborhood \( B_r(A) \) if similar matrices are inputs. Hence, there exists a differentiable function \( g \) satisfying the conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiability
In mathematics, "differentiability" refers to the ability of a function to have a derivative at each point in its domain. A function that is differentiable at a point has a gradient or a tangent plane that can be defined there. This concept is crucial when dealing with matrix functions, as smoothness in mapping is essential for many applications.

For a function to be differentiable, it must be continuous, and the limit of its difference quotient must exist as it approaches any point in its domain. In simpler terms, small changes in the input should lead to predictable and small changes in the output. In matrix theory, differentiability ensures that within a neighborhood of a matrix, there is consistency and smoothness in the behavior of the matrix function.
  • Continuity is necessary for differentiability but not sufficient on its own. A function can be continuous but not differentiable.
  • If a function is differentiable at a point, then softly varying inputs around this point should yield predictable variations in outputs.
Matrix Functions
Matrix functions extend traditional functions to matrices. This means we apply the concept of functions, like polynomials or exponentials, to matrices rather than to just numbers.

Matrix functions are interesting because they allow operations on matrices that mimic scalar functions. So, if you know how to compute a function like a square root or an exponential on numbers, you can extend this to matrices with some additional complexities.
  • Definition: A matrix function is a rule that assigns a matrix to another matrix, often written as \( f(A) \).
  • These functions often help solve differential equations or model systems in engineering and physics.
  • When extending functions to matrices, it's crucial to ensure definitions are consistent with matrix operations like addition and multiplication.
Neighborhood in Matrices
The concept of a "neighborhood" in matrix space is similar to neighborhoods in calculus but in the context of multi-dimensional matrix space. It defines a region around a particular matrix where certain properties, such as differentiability, are expected to hold.

In this exercise, the neighborhood is represented by \( B_r(A) \), an open set of matrices that are close to the matrix \( A \), within a specified radius \( r \). It means every matrix within this neighborhood satisfies certain conditions or properties, like having a differentiable function \( g \) applied to them.
  • Open neighborhoods are useful in analysis to explore the stability and changes in functions near a central point.
  • The idea is to have a sphere of matrices where similar operations yield similar results.
  • Matrix neighborhoods help isolate areas where approximations and continuity in functions are reliable.
Matrix Square Roots
In matrix theory, a "matrix square root" involves finding a matrix that, when squared, returns the original matrix. This is more complex than the scalar square root and generally involves more sophisticated methods.

The specific example in the exercise shows that given a matrix \( A \), there is a matrix \( g(A) \) that, when squared, equals \( A \). Computing the square root of a matrix is not straightforward due to their non-commutative nature. That means the order of multiplication affects the result, unlike scalar numbers.
  • Finding \( g(A) \) such that \( (g(A))^2 = A \) involves ensuring the consistent application of matrix multiplication rules.
  • Similar to scalar square roots, not all matrices possess square roots; existence depends on the matrix properties such as eigenvalues.
  • Matrix functions like this are crucial in fields that require decomposition or transformations of systems represented by matrices, such as signal processing.

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Most popular questions from this chapter

Show that if \(A\) is square, and \(\tilde{A}\) is what you get after row reducing \(A\) to echelon form, then either \(\tilde{A}\) is the identity, or the last row is a row of zeroes.

Find a number \(\epsilon>0\) such that the set of equations $$ \begin{aligned} x+y^{2} &=a \\ y+z^{2} &=b \text { has a unique solution near } 0 \text { when }|a|,|b|,|c|<\epsilon. \\ z+x^{2} &=c \end{aligned} $$

Show that any row operation can be undone by another row operation. Note the importance of the word "nonzero" in the algorithm for row reduction.

There are other plausible ways to measure matrices other than the length and the norm; for example, we could declare the size \(|A|\) of a matrix \(A\) to be the absolute value of its largest element. In this case, \(|A+B| \leq|A|+|B|\), but the statement \(|A \vec{x}| \leq|A||\vec{x}|\) is false. Find an \(\epsilon\) so that it is false for $$ A=\left[\begin{array}{ccc} 1 & 1 & 1+\epsilon \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right], \quad \text { and } \quad \overrightarrow{\mathbf{x}}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right]. $$

In this exercise, we will estimate how expensive it is to solve a system \(A \overrightarrow{\mathbf{x}}=\vec{b}\) of \(n\) equations in \(n\) unknowns, assuming that there is a unique solution, i.e., that \(A\) row reduces to the identity. In particular, we will see that partial row reduction and back substitution (to be defined below) is roughly a third cheaper than full row reduction. In the first part, we will show that the number of operations required to row reduce the augmented matrix \(\\{A \mid \vec{b}]\) is $$ R(n)=n^{3}+n^{2} / 2-n / 2. $$ (a) Compute \(R(1), R(2)\), and show that this formula is correct when \(n=1\) and \(2 .\) (b) Suppose that columns \(1, \ldots, k-1\) each contain a pivotal 1 , and that all other entries in those columns are 0 . Show that you will require another \((2 n-1)(n-k+1)\) operations for the same to be true of \(k\) (c) Show that $$ \sum_{k=1}^{n}(2 n-1)(n-k+1)=n^{3}+\frac{n^{2}}{2}-\frac{n}{2} $$ Now we will consider an alternative approach, in which we will do all the steps of row reduction, except that we do not make the entries above pivotal 1's be 0 . We end up with a matrix of the form at left, where \(*\) stands for terms which are whatever they are, usually nonzero. Putting the variables back in, when \(n=3\), our system of equations might be \(\begin{aligned} x+2 y-z &=2 \\ y-3 z &=-1 \\ z &=5, \quad \text { which can be solved by back substitution as follows: } \end{aligned}\) $$ z=5, \quad y=-1+3 z=14, \quad x=2-2 y+z=2-28+5=-21 $$ We will show that partial row reduction and back substitution takes $$ Q(n)=\frac{2}{3} n^{3}+\frac{3}{2} n^{2}-\frac{1}{6} n-1 \quad \text { operations. } $$ (d) Compute \(Q(1), Q(2), Q(3) .\) Show that \(Q(n)

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