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Show that if $x$ is in both $W$ and $W^{\perp},$ then $x=0$

The orthogonal compliment of the subspace $W$ is $W^{\perp}$ .Let the vector $x$ be in both $W$ and $W^{\perp}$ .Note that $x \in W^{1}$ if $x$ is orthogonal to every vector in a set that spans $W$ .As $x$ is also in $W$ , it must be orthogonal to itself. That is $x \cdot x=0$This happens only when $x=0$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 1

Inner Product, Length, and Orthogonality

Vectors

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This is problem number sixty one of this tour Capitalist states Addition Section two point three. It affects equals. This piece was function X squared. If X is irrational, zero effects is irrational. Prove that the Limited's experts zero f is equal to zero. We're going to approach this limit by understanding the F, as you can tell by the parts of it is a function that lines in between zero, which is its minimum and X squared, which is its maximum. So in this case, we will do our best to see if we can use eh squeezed hero to determine the limit as expert to zero for this function and with squeezed here, Um, what we do is we to recall we choose to function the one that is definitely always greater. Um around the region of exit protein A from this case zero. This function is always great enough and then the lower functions always less than half. And then we compare the limits and how they behave as limited has the element of it. Lim approaches zero for this function. Zero. That limit is equal to zero. The limit is export. Zero of this function X squared. It's also is here instant system. It must lie between these two values. Then we have confirmed. But with the squeeze serum, the Ltd's experts zero over their function in it must be equal to Syria.

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