In general, no. If A is a non-square mxn matrix, you have two cases:

1) If mn, then the image set of R^n in the mapping x \mapsto Ax is a proper subspace of R^m, and if you pick a point from the orthogonal complement of this subspace, you can't find the inverse image.

To better understand the idea, see the Rank-nullity theorem:

https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem

However, you can find the so-called pseudoinverse

https://en.wikipedia.org/wiki/Pseudoinverse

for any matrix by using the singular value decompostion (svd).

https://en.wikipedia.org/wiki/Singular_value_decomposition

It is not the actual inverse matrix, but the "best approximation" of such in the sense of least squares

https://en.wikipedia.org/wiki/Least_squares

Howver, if the actual inverse matrix exists, the pseudoinverse coincides with it. For example, in MATLAB you can find the pseudoinverse A by using the command pinv(A).

Remember the following properties: If A is m x n and the rank of A is equal to n, then A has a left inverse: an n-by-m matrix B such that BA = I. If A has rank m, then it has a right inverse: an n-by-m matrix B such that AB = I.

A m x n matrix is said to be full column rank if its columns are independent. This necessarily implies m >= n.

To find one left inverse of a matrix with independent columns A, we use the full QR decomposition of A to write

A = Q\left(\begin{array}{c}

R_1 \\

0

\end{array}

\right)

where R_1 is n x n upper triangular and invertible, while Q is m x m and orthogonal. So,

B = \left( \begin{array}{cc}

R_1^(-1) & 0

\end{array}

\right)

Similar procedure cold be accomplish in order to get the right inverse.

By other side, one way to compute the pseudo inverse (Moore–Penrose pseudoinverse) is by using the singular value decomposition, the QR method or the iterative method of Ben-Israel and Cohen

Hope the below wiki links will be useful for you

http://en.wikipedia.org/wiki/Invertible_matrix

http://en.wikipedia.org/wiki/Inverse_element#Matrices

Singular value decomposition. If M = U*S*V' (U = unitary, S = square diagonal of non-negative Reals, V = square unitary, and ' = notation for transpose), then:

For T = a certain diagonal matrix, V*T*U' is the inverse or pseudo-inverse, including the left & right cases. Specifically, the diagonal elements of T are the inverses of those of S, except that a 0 in S maps to a 0 in T.

Note that, conventionally, if M is non-square, then it is "tall" (#rows > #columns), and U has the same size as M. (If M is "wide" instead of "tall", then rework the analysis using the transpose of M.)

In general, no. If A is a non-square mxn matrix, you have two cases:

1) If mn, then the image set of R^n in the mapping x \mapsto Ax is a proper subspace of R^m, and if you pick a point from the orthogonal complement of this subspace, you can't find the inverse image.

To better understand the idea, see the Rank-nullity theorem:

https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem

However, you can find the so-called pseudoinverse

https://en.wikipedia.org/wiki/Pseudoinverse

for any matrix by using the singular value decompostion (svd).

https://en.wikipedia.org/wiki/Singular_value_decomposition

It is not the actual inverse matrix, but the "best approximation" of such in the sense of least squares

https://en.wikipedia.org/wiki/Least_squares

Howver, if the actual inverse matrix exists, the pseudoinverse coincides with it. For example, in MATLAB you can find the pseudoinverse A by using the command pinv(A).

Nandan, inverse of a matrix is related to notions of bijective, injective and surjective functions. That means you can invert a matrix only is it is square (bijective function). So a non singular matrix "must" not have an inverse matrix.

You can defined left (injective function) /right (surjective function) inverse for a non square matrix only if rank properties are satisfied and even though the left/right inverses are often not unique.

It is maybe just a matter of semantic to you but its implies plenty mathematical properties. For a matrix to be invertible it has to be square (non sufficient property of course). Otherwise you are referring to its"pseudo inverse". The "pseudo" is not just a meaningless word.

I support the answer of @Antti Rasila which is perfect.

In general you can't find it. Because of the conditions on the inverse.

Suppose A^{-1} is the inverse of an n x m matrix A. Then we must have that

A \cdot A^{-1} = A^{-1} \cdot A = I

where I is the identity matrix.

But if A is n x m, then if we can multiply by A^{-1} on both the left and the right,A^{-1} must be m x n. But then

A \cdot A^{-1} = I_n and

A^{-1} \cdot A = I_m.

But these must be equal, and hence n=m and A is square.

If the purpose of inverting the non-square matrix A is to solve a system of linear equations like Ax=B then you can multiply both sides of the matrix equation by the transpose of A so that it becomes (Transpose(A) A)X=Transpose(A)B. You can now invert Transpose (A) A and thus solve the system of equations.

The best inverse for the nonsquare or the square but singular matrix A would be the Moore-Penrose inverse. It is also a least-squares inverse as well as any ordinary generalized inverse. It becomes the regular inverse for a nonsingular matrix. It can be computed as follows:

Find the singular value decomposition of the mxn matrix as: A = P1ΔQ1T, where the rank of A is r, P1 is an mxr semiorthogonal matrix, Δ is an rxr diagonal matrix with positive diagonal elements called the singular values of A, and Q1 is an nxr semiorthogonal matrix. The Moore-Penrose inverse of A, denoted by A+ is the unique nxm matrix defined by: A = Q1Δ1P1T, where Δ1 is the inverse of Δ. It is the unique nxm matrix that satisfies the four conditions: (1) AA+A = A, (2) A+AA+ = A+, (3) (AA+)T = AA+, (4) (A+A)T = A+A.

In the note above, Q1T is the transpose of Q1 and P1T is the transpose P1.

Thanks everyone for enlightenment ...i got the concept behind pseudo-inverse of a non square matrix

From a practical point of view, QR factorization is the most efficient way to "invert" overdeterminated linear systems (ie matrices with a number of lines larger then the number of columns).

In fact as it was writen fore, for a matrix A of size m×n with m>n, solving the problem

Ax=b (1)

is a least square problem.

The matrix A can be factorized as the product of an orthogonal matrix Q (m×n) and an upper triangular matrix R (n×n), thus, solving (1) is equivalent to solve

Rx = Q^T b

For more detail, refer to the book of Golub et al. "Matrix computation".

To find the inverse of non-square matrix by using generalized inverses :  Moore-Penrose inverse and it has so many representations of this inverse ( SVD representation, Integral representation,, ....etc) and also can be easily  computed which is depends on the rank of the given matrix. If you are interested , please go and search about my name : zeyad al-zhour in Google and then you will find some of my published papers in this topic.

I'd present the situation as follows, as would be applied to the case of exploring the convolution coefficients in the case of Savitzky-Golay filters:

You have the 2M+1 sample points in a given (Gaussian) spectrum assembled in the form of a (2M+1)*1 column-vector, i.e. \hat{x}=(x_{-M}, x_{-M+1},..., x_0, x_{M-1}, x_M})^T.

The coefficients of a polynomial expansion of these 2M+1 data points are themselves in an N+1 dimension column vector (where N is the order of the polynomial fit) according to: \hat{a}=(a_0,a_1,...,a_N)^T. In general N$\le$2M and so the matrix pre-multiplying \hat{x} to get \hat{a} is not square. The relation between \hat{x} and \hat{a} is given by  \hat{x}=A\hat{a}, where A is a non-square matrix containing integer powers at all positions.

To get \hat{a}, you must firstly multiply each side of the above equation by A^T, the transpose of A giving A^T*\hat{x}=A^T*A*\hat{a}, the product matrix A^T*A is itself square since it involves the multiplication of a non-square matrix by its transpose: (n*m)*(m*n)=(n*n) where in the first matrix n is the number of rows and m is the number of columns and vice versa for the second.

Thus, \hat{a}=(A^T*A)^{-1}A^T*\hat{x}, since the inverse of a square matrix is more straightforward to calculate. I thought the particular example together with the mathematical subtleties makes the explanation more rigid for the mind to digest.

Cholescy factorizaiton can used for non square matrixs.

does anyone know how to solve xA=b? post multiplying by pinv(A) does not work...

My suggestion is to transpose the equation; in your problem, x is a row vector and becomes a column vector under transposition. The transposed equation looks like A'x'=b', where A' is the transpose of A. This is easily solved using whatever resources we have at present.

• First of all we should know whether the system is consistent or not. The solution may be given by x=b X, where X is pseudo inverse of A ( i.e.,a matrix X satisfying AXA = A).  Now bXA = xAXA = xA = b (since xA=b). If it is inconsistent system then go for least squares pseudo inverse.

There are many good answers here.

I think reading in Wikipedia

https://en.wikipedia.org/wiki/Generalized_inverse

may give a more complete answer.

Hope it helps.

Itzhak

Wikipedia answer is almost complete but fails to mention the least squares inverse AL that satisfies conditions (1) A(AL)A = A and (3) Tranpose(A(AL)) = A(AL) of the Moore-Penrose inverse. This invrse is quite useful in statistics. For more information on generalized inverses, see Matrix Anlysis for Statistics by James R. Schott.