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Hello, today we're going to prove that if a and b are end -by -on matrices and a is invertible, then the nullity of ab equals the nullity of b, and that equals the nullity of b.
00:14
First, we're going to prove that the null space of b is a subset, the null space of ab.
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So, null space of b is a subset of a sub set of a -b.
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So null space of b is a subset of ab.
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And we're going to define x.
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As the null space.
00:50
Therefore, if we have ab times x equals 0, then we can take this, multiply it by using red a inverse.
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And because a is invertible, we know that a times a inverse equals the identity matrix.
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So we have the identity matrix.
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Times b equals 0.
01:35
And i think times the identity matrix of itself, so b times x equals 0.
01:41
And this tells us that b is indeed a subset of ab.
01:47
Next, we want to do sort of the reverse and show that the null space of ab is contained within b.
01:54
So let's do b .s times our null space equals 0 .0 .0.
02:05
So on the converse side, we can multiply both sides by a.
02:14
Let's use red for that.
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Go back a little bit.
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Multiply by a.
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So this gives us ab x equals zero.
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So this tells us that the null space of ab is contained within b.
02:33
Therefore, therefore, ab and b have the same null space.
02:50
Or in other words the nullity of ab equals the nullity of b.
03:17
So now let's go down and use our rank nullity theorem to try to investigate this a little deeper.
03:27
So we have our rank nullity theorem using blue.
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So if we do this for the matrix ab, the rank of ab, plus the nullity of ab equals the number of columns of ab.
04:11
And similarly, the rank of b plus the nullity of b equals the number of columns of b.
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But we also know that the rank of ab, going back over to the left equation, the rank of ab is the same thing as writing the dimension of the column space of ab.
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And that the nullity is the same as the dimension of the null space of the ab.
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And that's the same is true for b.
05:02
So we can therefore rewrite what we have in terms of the nullity of ab...