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Suppose $A$ is invertible and orthogonally diagonalizable. Explain why $A^{-1}$ is also orthogonally diagonalizable.

We can conclude that $A^{-1}$ is an orthogonally diagonalizable matrix.

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 1

Diagonalization of Symmetric Matrices

Introduction to Matrices

McMaster University

Baylor University

Lectures

01:32

In mathematics, the absolu…

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$A$ is a $3 \times 3$ matr…

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In Exercises 31–36, mentio…

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Since we know a IHS or thermally diagonal Izabal, we could write a s a equals p D Ping furs where p is an orthogonal matrix and D is a diagonal matrix. And we also know that a Ihsan vertebral so we can take the inverse off both sides of the equation. So now we have day in vers equals the embers off PDP fingers and using the property of fingers we're basically distributing the inverse, um, into the expression between the parentheses and reverse the expression. So now we have thean vers off paying vers times Demers times pain givers. And now, um Thean vers of painters is just p itself. So now we have the expression right? This also we know that b a d c diagnose matrix. So being vers off a diagonal matrix, it's still a dive matrix. We can call it be and then the expression looks like a members vehicles P B papers. So, according to the definition of a or thoughtfully Diagne allies, full matrix am various s A dyke? No, or talking really dying, No sizeable matrix. Um we can also prove this part. So since the is a diagnosed matrix, we know that D equals the transport. We want to prove that Denver's it's also a diagonal matrix. You can take the inverse off both sides like this, um, using the property of transposed and, um, favours this part. Also, you calls Devers the transports off Demers. And now we have Denver's equals the transport of Dave Demers. Um, so using the definition off, um, a diagnose matrix, Dave Ear's is also a diagnosed matrix. So now we proved this part, and we completed a whole proof.

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