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Suppose you have a model of the form:
$$X \beta= Y$$ where X is a normal 2-D matrix, for ease of visualisation. Now, if the matrix $X$ is square and invertible, then getting $\beta$ is trivial: $$\beta= X^{-1}Y$$ And that would be the end of it.

If this is not the case, to get $\beta$ you’ll have to find a way to “approximate” the result of an inverse matrix. $X^\dagger = (X'X)^{-1}X'$ is called the (left)-pseudoinverse, and it has some nice properties that make it useful for this application.

In particular, it is unique, and $XX^\dagger X=X$, so it kind of works like an inverse matrix would $(XX^{-1}X = XI = X)$. Also, for an invertible and square matrix (i.e. if the inverse matrix exists), it is equal to $X^{-1}$.

Also it gets the shape of the matrix right: If $X$ has order $n \times m$, our pseudoinverse should be $m \times n$ so we can multiply it with $Y$. This is achieved by multiplying $(X'X)^{-1}$, which is square $(m \times m)$, with X' $(m \times n)$.

Suppose you have a model of the form:
$$X \beta= Y$$ where X is a normal 2-D matrix, for ease of visualisation. Now, if the matrix $X$ is square and invertible, then getting $\beta$ is trivial: $$\beta= X^{-1}Y$$ And that would be the end of it.

If this is not the case, to get $\beta$ you’ll have to find a way to “approximate” the result of an inverse matrix. $X^\dagger = (X'X)^{-1}X'$ is called the (left)-pseudoinverse, and it has some nice properties that make it useful for this application.

In particular, it is unique, and $XX^\dagger X=X$, so it kind of works like an inverse matrix would $(XX^{-1}X = XI = X)$. Also, for an invertible and square matrix (i.e. if the inverse matrix exists), it is equal to $X^{-1}$.

Suppose you have a model of the form:
$$X \beta= Y$$ where X is a normal 2-D matrix, for ease of visualisation. Now, if the matrix $X$ is square and invertible, then getting $\beta$ is trivial: $$\beta= X^{-1}Y$$ And that would be the end of it.

If this is not the case, to get $\beta$ you’ll have to find a way to “approximate” the result of an inverse matrix. $X^\dagger = (X'X)^{-1}X'$ is called the (left)-pseudoinverse, and it has some nice properties that make it useful for this application.

In particular, it is unique, and $XX^\dagger X=X$, so it kind of works like an inverse matrix would $(XX^{-1}X = XI = X)$. Also, for an invertible and square matrix (i.e. if the inverse matrix exists), it is equal to $X^{-1}$.

Also it gets the shape of the matrix right: If $X$ has order $n \times m$, our pseudoinverse should be $m \times n$ so we can multiply it with $Y$. This is achieved by multiplying $(X'X)^{-1}$, which is square $(m \times m)$, with X' $(m \times n)$.

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Suppose you have a model of the form:
$$X \beta= Y$$ where X is a normal 2-D matrix, for ease of visualisation. Now, if the matrix $X$ is square and invertible, then getting $\beta$ is trivial: $$\beta= X^{-1}Y$$ And that would be the end of it.

If this is not the case, to get $\beta$ you’ll have to find a way to “approximate” the result of an inverse matrix. $X^\dagger = (X'X)^{-1}X'$ is called the (left)-pseudoinverse, and it has some nice properties that make it useful for this application.

In particular, it is unique, and $XX^\dagger X=X$, so it kind of works like an inverse matrix would $(XX^{-1}X = XI = X)$. Also, for an invertible and square matrix (i.e. if the inverse matrix exists), it is equal to $X^{-1}$.