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Prove that the set of invertible real \(2 \times 2\) matrices is dense in the set of all real \(2 \times 2\) matrices.
in Mathematics by Platinum (164,226 points) | 333 views

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To prove that the set of invertible real \(2 \times 2\) matrices is dense in the set of all real \(2 \times 2\) matrices, we will show that for any real \(2 \times 2\) matrix \(A\), there exists a sequence of invertible real \(2 \times 2\) matrices \(B_n\) such that the sequence \(B_n\) converges to \(A\) as \(n\) goes to infinity.

To construct the sequence \(B_n\), we will use the fact that the determinant is a continuous function on the set of all \(2 \times 2\) matrices. Specifically, let det be the determinant function, which maps a \(2 \times 2\) matrix to its determinant. Then for any real \(2 \times 2\) matrix \(A\), we can define a sequence \(B_n\) as follows:

\[B_n = (1 - \frac{1}{n}) A + \frac{1}{n} I\]

where \(I\) is the \(2 \times 2\) identity matrix. Then

\[det(B_n) = (1 - \frac{1}{n})\]

\[det(A) + \frac{1}{n} det(I) \]

\[= 1 - \frac{1}{n} + \frac{1}{n} = 1\]

for all n, so \(B_n\) is invertible for all \(n\).

Now, we will show that the sequence \(B_n\) converges to \(A\) as \(n\) goes to infinity. To do this, we will use the fact that the matrix norm is a continuous function on the set of all \(2 \times 2\) matrices. Specifically, let || || be a matrix norm, which maps a \(2 \times 2\) matrix to a nonnegative real number. Then

\[||B_n - A|| = ||(1 - 1/n) A - (1 - 1/n) A + (1/n) I|| \]

\[= ||(1/n) I|| = 1/n\]

for all \(n\), so the sequence \(B_n\) converges to \(A\) as \(n\) goes to infinity. This completes the proof.
by Platinum (164,226 points)

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MathsGee Android Q&A

MathsGee Android Q&A