We know from Table 1 that similar matrices have the same...

We know from Table 1 that similar matrices have the same rank. Show that the converse is false by showing that the matrices $$A=\left[\begin{array}{ll}1 & 0 \\0 & 0\end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right]$$ have the same rank but are not similar. [Suggestion: If they were similar, then there would be an invertible $2 \times 2$ matrix $P$ for which $A P=P B .$ Show that there is no such matrix.]

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00:01 Okay, so the question tells us that similar matrices have the same rank, and we want to show that the converse is false, i .e.

00:08 That there are matrices of the same rank, which are not similar.

00:12 So here, these matrices a and b, have the same rank.

00:16 So the dimension of their column space is the same, or their image.

00:21 So it's just one in both cases, so they have the same rank.

00:25 But we're going to show that they're not similar.

00:27 And the hint is that if they were similar, we would have an invidable 2x2 matrix p such that a p is equal to p times b and we're going to show that if we try and do this we come to a contradiction so basically we're just going to define p to be a 2 by 2 matrix a b c d if we do this then a times p well a is 1 -0 -0 times p which is ab c so this is a, p.

01:04 Okay, so what we're going to get is the following.

01:07 The first entry will be 1 times a plus 0 times c, which is just a.

01:11 The next entry will be 1 times b plus 0 times d, which is just b.

01:17 The next entry is going to be 0 times a plus 0 times c, which is 0.

01:22 And the next entry will be 0 times b plus 0 times d, which is 0.

01:27 So this is ap.

01:29 Now let's look at p b.

01:33 Well, p is a, b, c, d, and b is 0, 1 ,0...

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